@Jim. If p does not divide the order of the group, a f.g. $\mathbb Z_p G$ module is projective iff the underlying $\mathbb Z_p$-module is. That the condition is necessary is clear. That it is sufficient follows eg from the spectral sequence $ H^i(G,Ext^j_{Z}(M,N)) => Ext^{i+j}_{ZG}(M,N)$ since $G$ has no cohomology on $\mathbb Z_p G$-modules (invariants are the image of the usual idempotent). By the way, the assumption that $M$ is a free $\ZM$-module in the CR reference above corresponds to the "extended" assumption on localisations at all primes in the first sentence of my answer

Ah OK, I have just discovered the "show 2 more comments" button and I now understand that you got it before my answer. Well, in any case I don't care about reputation, so you should give it to Piotr. For that, he should write up an official "answer" I guess, so you can validate it as THE answer.