That's a pretty conjecture.  Does it extend to higher powers of 5 in any obvious fashion?

If I should gamble I would guess that

${F_{{5^m}n}}(q)$ is divisible by ${[{5^m}]_{{q^n}}}$. 

In my experience, proving things like this, it's often easier to prove a stronger statement than a weaker one.  Some of the techniques in my paper Factors of Sums of Powers of Binomial Coefficients might be relevant here.

The first part of your original conjecture is true.  It can be proved as follows: 

take a vector of consecutive q-Fibonacci values 

u_n = F_n

      F_{n-1}

and let A_n be the matrix with rows [1 , q^{n} ] [ 1, 0], so that u_n satisfies the recurrence u_n=A_{n-2} u_{n-1}.

Now, since we are working mod [5]_q, we can replace q by a fifth root of unity,

and we find that A_0 A_1 A_2 A_3 A_4 is a matrix with a 0 in the lower left corner, implying that F_{5n} is congruent to 0 mod [5]_6.

It appears that if we let q be a root of the 5^k th cyclotomic polynomial, then the product of the first 5^k A's is very nice.  This may imply your strong conjecture.

Indeed, if we let C be the product of the first 5^k A's, then it appears that  C has entries 

 [  q^{3*5^{k-1}} + q^{2*5^{k-1}} + 1     -2*q^{3*5^{k-1}} - 2*q^{2*5^{k-1}} - 1  ]

[       0                                                        -( q^{3*5^{k-1}} + q^{2*5^{k-1}} )         ]

As of now, this is purely empirical: it holds for k up to 4.

And of course, the result for 5^k follows from the result for 5, since the cyclotomic polynomial for 5^k is just the same as the one for 5 with x replaced by x^{5^{k-1}}.

@Neil J. Calkin: Thank you very much for the reference to your paper.

I think your proof with matrices is equivalent to the proof which I have given for the congruence mod [5]_q.

${F_{{5^k}n}}(q) \equiv 0\bmod {[{5^k}]_q}$ is equivalent with ${F_{{5^k}}}(q) \equiv 0\bmod {[{5^k}]_q}$ 

because ${q^{{5^k} + n}} \equiv {q^{{5^k}}}\bmod {[{5^k}]_q}$ and therefore  ${F_{{5^k}(n + 1)}}(q) \equiv {F_{{5^k}n}}(q){F_{{5^k}n + 1}}(q) \equiv 0\bmod {[{5^k}]_q}.$

Therefore it suffices to show that

${F_{{5^k}}}(q) \equiv 0\bmod {[{5^k}]_q}.$