I think it can be found the following way. If R is a commutative local ring then every idempotent matrix is ~~equivalent~~ conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is projective is free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank r diagonal idempotent under the conjugation action of GL(R) for $R=Z_{p^{k}}$.

**Added.** I believe the stabilizer of the rank r idempotent for a local ring is GL(R,r)xGL(R,n-r)$ like in the field case and so a formula is easily found.

**Added.** I compute the answer for nxn matrices over $Z_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|Gl(n,p)|}{|Gl(r,p)|Gl(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the Gl(m,p) are of course well known.