The answer is 'yes' when your group $G$ is word-hyperbolic. This can be deduced from Sageev's theorem. I'll explain this here, but a good reference is Hruska--Wise's paper 'Finiteness properties of cubulated groups' (arXiv:1209.1074v2).
The first step is to notice that $G$ admits a proper, but not cocompact, action on a cube complex.
Lemma: Suppose $|G:N|<\infty$ and $N$ acts properly on a CAT(0) cube complex $X$. Then $G$ acts properly on $X^{|G:N|}$.
The proof of this is the usual messing around with the induced representation, so I'll leave it as an exercise. The one extra fact about this action that we will need is that the new hyperplane stabilizers are precisely the $G$-conjugates of the hyperplane stabilizers in $N$.
When $G$ is word-hyperbolic, the Schwarz--Milnor Lemma implies that $X$ is Gromov-hyperbolic. We also have the following facts about hyperplane stabilizers in $G$.
Hyperplane stabilizers in $N$ are codimension-one subgroups. Since this is a coarse property, the same is true of hyperplane stabilizers in $G$.
Hyperplanes in $X$ are convex. Since quasigeodesics in Gromov-hyperbolic spaces are uniformly close to geodesics, it follows that hyperplane stabilizers are quasiconvex in $N$, and hence in $G$.
Associated to any finite collection $\lbrace H_i\rbrace$ of codimension-one subgroups in a group $G$, Sageev constructed a CAT(0) cube complex on which $G$ acts by isometries. In this case, we will take $\{H_i\}$ to be a set of $G$-conjugacy representatives for the hyperplane stabilizers in $G$.
Sageev's theorem: If the $\{H_i\}$ are all quasiconvex then $G$ acts cocompactly on the associated cube complex.
By 1 and 2 above, Sageev's theorem applies. It remains to prove that $G$ also acts properly. This is a matter of making sure that we have chosen a large enough collection of hyperplane stabilizers.
The Hruska--Wise paper contains some useful criteria for checking that the action is proper. Roughly speaking, their Theorem 5.4 says that, as long as the axis of every infinite-order element crosses some hyperplane, the action is proper. But, indeed, if $g\in G$ has infinite order then it shares an axis with $g^{|G:N|}\in N$, and this certainly crosses a hyperplane, since the action of $N$ on $X$ was proper.
This completes the proof in the word-hyperbolic case (modulo filling in some details).
Finally, I'll just mention that much of the purpose of the Hruska--Wise paper is to generalize these ideas to the relatively hyperbolic setting. In particular, something similar should be true there, subject to suitable restrictions on the action of the parabolic subgroups.
