Yes, this is true. It suffices for $X$ to have the homotopy type of a CW-complex (this is true of smooth manifolds; see e.g. here or here).
I'm going to assume that you're using a definition of $EG$ that includes something like: $EG$ is a $G$-CW-complex, so that it is built by iteratively taking pushouts of diagrams of the form $$ D^n \times G \leftarrow S^{n-1} \times G \rightarrow Y. $$ As a result, the space $EG \times_G M$ is formed by an iterated sequence of pushouts $$ D^n \times M \leftarrow S^{n-1} \times M \rightarrow Z. $$ (This comes with the standard warnings about probably having to use compactly generated spaces so that products, quotients, and the direct limit topology interact well.)
Each of these pushouts is the mapping cone of the map $S^{n-1} \times M \to Z$. This mapping cone would not necessarily be a CW-complex even if $M$ and $Y$ were (the map would have to be cellular for that), but if $M$ and $Z$ both have the homotopy type of CW-complexes, the cone does have the homotopy type of a CW-complex (it's homotopy equivalent to a cellular map, and that equivalence carries across to an equivalence on mapping cones). By inducting on the cell structure of $EG$, we can assume that $Z$ has the homotopy type of a CW-complex and find that this next pushout map $Z \to Z'$ is homotopy equivalent to a cell inclusion of CW-complexes. Taking colimits we get the desired result.
