The answer might be negative. Let me explain. An integrable peakon equation usually has the following structure (you can refer to some papers written by Fokas/Fuchssteiner/Olver around 1996): 

u_{t}=P \cdot (\partial_x-\partial_{x}^{3})^{-1} u_x

where P is a Hamiltonian operator (also known as implectic operator). The term  (\partial_x-\partial_{x}^{3})^{-1} u_x is crucial for admitting peakon, therefore the operator (\partial_x-\partial_{x}^{3})^{-1} should be remained (can not be absorbed into P). Then one has to introduce u=(1- \partial_x^{2})m so as to get rid of nonlocal terms. Thus, the equation is non-evolution in terms of m.

There might be exceptional, but I do not know.

The answer can be positive. Indeed, Camassa-Holm equation can be considered as a (local) evolution equation by taking inverse of the operator 1-\partial_x^2. Note that (1-\partial_x^2)^{-1}=1+\partial_x^2+\partial_x^4+\dots. But, of course, as an evolution equation, Camassa-Holm contains infinitely many terms (each term is a differential polynomial in u).

I have finally discovered an answer to my question and it is positive.  An example of an evolution equation admiting peakon solutions can be found in the link below.

Article A family of wave equations with some remarkable properties