This problem is answered in the literature, with a caveat.
As indicated in the comments, it follows from the orbifold theorem + geometrization conjecture (to handle the case of orbifolds without fixed points).

The caveat is that the action needs to be assumed smooth (or PL). Otherwise, there exists wild involutions such as the Bing involution. One could incorporate this into an action on $\mathbb{R}^3$ by taking an orbifold quotient whose underlying space is $S^3$ and admits a reflection symmetry, then take a Bing involution preserving the orbifold locus by inserting into a part of the sphere being reflected. The resulting group action would have quotient which is not an orbifold. I think this was an underlying assumption of the problem as stated, but I thought I would clarify.

With the smoothness assumption, now this follows from the orbifold theorem (see Problem 3.46 of Kirby’s list posed by Geoff Mess). I won’t go through the history of the proofs of cases of this (see the link), but it follows for orientable orbifolds modulo previous results by the proof of the geometrization theorem by Perelman.

For the non-orientable case, take an index-two subgroup $G’<G$ which is orientation-preserving. By the orientable case (and no $Z+Z$ subgroup assumption), $\mathbb{R}^3/G’$ is an orientable hyperbolic 3-orbifold. By Selberg’s lemma, there is a finite-index subgroup $G’’ < G’$ for which $M’’=\mathbb{R}^3/G’’$ is a manifold. By passing to a further subgroup (the *core*), we may assume that $G’’\lhd G$, so $G/G’’$ acts as a finite group of transformations of $M’’$. Now one may apply a result of Dinkelbach-Leeb (Theorem H) to see that the quotient orbifold $M’’/(G/G’’)= \mathbb{R}^3/G$ is a hyperbolic orbifold.

topologicalorbifold, not a smooth one. Do you still see how to put a $G$-invariant hyperbolic metric on $M$? $\endgroup$