As Najib says in the comments to the question, the proof of the classical statement can be easily-ish adapted to the equivariant case. Let's see the details
Lemma Let $X$ be a $G$-CW-complex and let $f:G/H\times S^n\to X$ be a $G$-equivariant continuous map. Then $f$ factors up to equivariant homotopy through the $n$-skeleton of $X$
Proof: I'm goint to sketch the proof only for when $G$ is a finite group. When $G$ is a compact Lie group this is just a special case of the cellular approximation theorem (theorem I.3.4 in May's Equivariant Homotopy and Cohomology Theory). A $G$-equivariant continuous map from $G/H\times S^n$ is the same thing as a continuous map from $S^n$ to $X^H$. Since the cellular filtration on $X^H$ is the restriction of the one for $X$, we can consider only the case where $G$ is trivial. But this is well known. $\square$
Armed with the lemma, we can prove that every $G$-complex $Y$ is homotopy equivalent to a $G$-CW-complex. I will prove only the case where $Y$ has finitely many cells, the general case just needs a transfinite induction that I don't want to write down right now.
We will proceed by induction on the number of cells. If $Y$ has only one cell, then it has of course the structure of a $G$-CW-complex, obtained by using a CW structure on the disc.
Suppose now that the thesis is true for all $G$-complexes with $m$ cells. Then, if $Y$ is a $G$-complex with $(m+1)$ cells, we can write it as the pushout $Y=Y'\amalg_{G/H\times S^n} G/H\times D^{n+1}$ for some $H$ and $n$. But, by the previous lemma, the attaching map $G/H\times S^n\to Y'$ factors up to homotopy through the $n$-skeleton of $Y'$. So we can attach it together with the other $(n+1)$-cells without changing the homotopy type of $Y$. Hence $Y$ has the homotopy type of a $G$-CW-complex.