as a matter of fact, natural TiO2 is always bluish. This is because natural TiO2 is always contains considerable oxygen vacancies that result reduced TiO2. The vacancy state turn as Ti3+, and creates an in-gap state and form an intermediate band  just below the conduction band of the host material. You are using a high RPM that raises temperature and promote the reduction of TiO2. I think using pressurized oxygen ambient to inhibit reduction during your ball milling is a possible option. 

Difficult to tell without knowing further details...

The blue color is probably due to some charge transfer Ti 3+ -Ti4+ and the presence of oxygen vacancies. In that case youo will not be able to find impurities because the color comes from the material itself.

Do you mill the powder dry or in a solvent ? Does the titania powder contain any organic additives ? Titania is relatively easy to reduce if the temperature gets high enough and there are some reducing agents (traces of organic) present.

Maybe you should anneal your powder gefore at 400°C for 2h to remove all organic contaminations, if the blue color still appears then anneal the milled product.

Probably the blue color will disappear if you anneal the milled powder at moderate temperature.

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regarding Ti3+ surface defect properties, see

http://www.hindawi.com/journals/jnm/2012/831524/

(see section 3.2 and the references for its relation to the blue coloration)

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Please try to use XRD to see if there is any change in structure, and XPS to see if there is any change in oxidation state of Ti.

I am milling the powder through dry prcess and I do not use any solvent in the process.

Ti is a transition metal and has various oxides, e.g., TiO2 (IV oxide) , Ti2O3 (III), TiO. only Valace IV oxide is non absorbing and the other has color. In your process you are redusing TiO2 to TiOx where x

Are you sure it turns "blue" but not "light grey"? I guess it can be more reasonable if the milled TiO2 turns light grey from white. There should be many OH groups on the surface of TiO2 for the sake of stability. When milled in a high-energy and dry process, TiO2 can partially lose OH groups to form oxygen vacancies. This leads the color change. FTIR should be useful to detect that.

Is there any possibility to reverse the change occuring if I anneal the powder at 600 C

An EDX analysis can be used to ascertain presence of zirconia as an impurity in ball milled product. I experienced contamination of zirconia in some of the ball milled products. In case of zirconia coontamination your product should turn " light grey" and the colour becomes more intense with increasing milling duration (multiple hours).

If the color change is from reduction (formation of oxygen vacancies) then calcination in air should restore the color.  If it is contamination (incorporation of zirconia or other impurity) then it is unlikely that the color be eliminated by calcination.

As the case of sio2 .its color turn blue when  reacts with oxygen.same may be the case with ti

you should decide weather it is only a morphology effect (multiple reflections due to changes in surface changes) or a change in the mterial, as the answers above already mentioned.

So:

- for morphology changes: AFM and roughness measurements 

- for changes in material: EDX (contamination - the analysis is not very reliable), XPS (oxygen vacancies and non-stoichiometry) or EPR or RAMAN

Please used XRD+EDX to figure out the phases present and whether any entrapment of other particles to turn TiO2 into blue.

CONTAMINATION LEVEL can be sorted out from EDX technique.

Ideal ball to powder weight ration is :10:1

I have prepared the titania nanoparticles in mechanical grinding method but I did not see such colour change. For details refer: https://www.researchgate.net/publication/244990066_Titanium_dioxide_%28TiO2%29_Nanoparticles_XRD_Analyses_An_Insight

Take a XRD of the grinded material and confirm, is there any contamination or not?

Article Titanium dioxide (TiO2) Nanoparticles XRD Analyses: An Insight

as a matter of fact, natural TiO2 is always bluish. This is because natural TiO2 is always contains considerable oxygen vacancies that result reduced TiO2. The vacancy state turn as Ti3+, and creates an in-gap state and form an intermediate band  just below the conduction band of the host material. You are using a high RPM that raises temperature and promote the reduction of TiO2. I think using pressurized oxygen ambient to inhibit reduction during your ball milling is a possible option. 

Formation Mechanism of Ti3+ Surface Defects. TSD can be generated by reduction of Ti4+ ions. There are two typic processes for Ti4+ reduction to Ti3+ .TSD lie in two aspects. One is that Ti3+ ions are the origin of the blue coloration or coloration center and the other is that Ti3+ species has a characteristic visible absorption spectrum.

TiO2 can easily become blueish. -> Liu lu's and A. Ouerdane comments. Annealing at 300 degrees Celsius at air works.

I use acetone when milling delicate compounds (mainly water-soluble and somewhat air-soluble sulfates), but of course we should be aware of not using acetone in contact with plastic. Btw, regarding the blue colour, anatase (one of the natural TiO2 forms) is quite often gray-blue.

Rutile becomes blue on reduction BUT interestingly neither lunar nor Allende meteorite rutile exhibits a color change although Ti4+ to Ti3+ [oxygen-deficient Magneli phases] is demonstrated.

I agree with Prof. Liu lu .

Titania becomes blue on reduction. It exhibits a color change although Ti4+ to Ti3+ as Stephen Haggerty said .

I agree with Fatma Zohra Benabid

I agree with Prof Liu Lu.

The colour change is due to possible reduction of Ti4+ to Ti3+ since the conditions of the reaction favours the reduction as also stated by Prof. Liu Lu

This is typical in MB with different applications for example extrusion or injection products. There are MB 2 types of Mb white that have a blue and neutral color, due to the particle siz.

If the particle size is small (< 300 nm) in suspension the finest particles will have a bluish tinge due to Rayleigh scattering.