Whenever $G$ is a finite group, a $G$-CW-complex structure on $X$ is equivalent to a CW-complex structure on which $G$:
(1) The image of an open cell is another open cell (and boundaries are preserved).
(2) Fixed cells are point-wise fixed.
This shows that whenever $X$ admits a finite (locally finite, etc.) $G$-CW-complex structure, then it also admits a finite (resp. locally finite, etc.) CW-complex structure.
I was wondering about the converse:
If $X$ is a $G$-space with the structure of a finite (locally finite, etc.) CW-complex, can it be turned into a finite $G$-CW complex without changing the homotopy properties of the action (a new $G$-CW complex $Y$ which is $G$-homotopy equivalent to $X$ is allowed)?
I know the answer is no for arbitrary groups, as P. May mention in Equivariant Homotopy and Cohomology Theory with the example of topological $G$-manifolds. But maybe the finiteness assumption for the group is enough.
Even if the approximation theorem allows replacing the original action by a cellular action, I do not know how to overcome the second condition.
Note. Although any $G$-space can be approxitamed by a $G$-CW complex, this approximation, even for a finite group $G$, is wild, at least as explained in P. May I Theorem 3.6, for it is indexed in elements of the homotopy groups.
EDIT.
I think my question can be reformulated as follows:
Let $X$ be a $G$-space with the homotopy type of a finite (resp. locally finite, etc.) CW-complex. As a $G$-space, does $X$ have the homotopy type of a finite (resp. locally finite, etc.) $G$-CW complex?
Let $G$ be a cyclic group of prime order $p$. Then $EG$ has the homotopy type of a finite CW-complex, since it is contractible, but it cannot be equivariantly homotopy equivalent to a finite $G$-CW-complex, since this would imply that $BG=EG/G$ is a finite CW-complex, but $H_\ast(BG;\mathbb{F}_p)$ is non-trivial in every dimension.
The above argument actually works for any finite group $G$, that is $H_\ast(BG)$ is non-trivial in infinitely many dimensions, but this seems to be more difficult to prove, see this Mathoverflow post.