Actually, having finitely many orbits of hyperplanes is quite common. For instance:

**Proposition:** (Sageev) Let $G$ be a finitely generated group acting on a CAT(0) cube complex $X$. Then there exists a $G$-invariant convex subcomplex $Y \subset X$ containing only finitely many $G$-orbits of hyperplanes.

*Sketch of proof.* Let $s_1, \ldots, s_n$ be generators of $G$ and $x_0 \in X$ a vertex. Set $Y$ as the convex hull of the orbit $G \cdot x_0$. The hyperplanes of $Y$ are exactly the hyperplanes of $X$ separating two vertices of $G \cdot x_0$. Let $J$ be such a hyperplane. So there exist $g,h \in G$ such that $J$ separates $gx_0$ and $hx_0$. By translating, we may suppose that $g=1$. Write $h$ as a word of generators $r_1 \cdots r_k$. By looking at a path
$$[x_0,r_1x_0] \cup [r_1x_0,r_1r_2 x_0] \cup \cdots \cup [r_1 \cdots r_{k-1}x_0, r_1 \cdots r_{k-1}r_k x_0]$$
from $x_0$ and $hx_0$, we find that there exists some $i$ such that $J$ separates $r_1 \cdots r_ix_0$ and $r_1 \cdots r_ir_{i+1} x_0$. Up to translating $J$, we may suppose that $J$ separates $x_0$ and $r_{i+1}x_0$.

In other words, any hyperplane of $Y$ has a translate which separates $x_0$ from $s x_0$ for some generator $s$. It follows that there exist only finitely many orbits of hyperplanes in $Y$. $\square$

In other words, if we are looking at finitely generated groups only, we can always suppose that the cube complex contains only finitely many orbits of hyperplanes, without any assumption on the action.

allstabilizers are trivial, i.e. the set of orbits such that the stabilizers are non-trivial isempty, quite the opposite of being infinite. (Of course, the stabilizers may all be trivial in more elaborate examples, such as YCor's.) $\endgroup$ – Victor Protsak Aug 2 '19 at 13:46