As pointed out in the comments, the answer is negative. But we can prove even more:

**Proposition:** Let $G$ be a group acting properly and cocompactly on a CAT(0) cube complex $X$. Then there exist a $G$-invariant convex subcomplex $Y \subset X$ and an isometry $g \in G$ which not parallel to any of the hyperplanes of $Y$.

In other words, the question has a negative answer in most of the reasonnable situations.

*Sketch of proof.* Let $Y \subset X$ be a convex subcomplex on which $G$ acts essentially [CS, Proposition 3.5]. Decompose $Y$ as a Cartesian product $Y_1 \times \cdots \times Y_n$ of irreducible cube complexes and fix a finite-index subgroup $\dot{G} \leq G$ which preserves the product structure. Now look at the induced (cocompact) action $\dot{G} \curvearrowright Y_i$. As a consequence of [CS, Theorem 6.3], $\dot{G}$ contains a strongly contracting isometry of $Y_i$ (i.e., an isometry which skewers a pair of strongly separated hyperplanes).

Thus, $\dot{G}$ acts simultaneously on $n$ hyperbolic graphs, namely the contact graphs of $Y_1, \ldots, Y_n$, and $\dot{G}$ contains a loxodromic isometry for each action. By applying [CU], it follows that $\dot{G}$ contains an element inducing a loxodromic isometry simultaneously in all the contact graphs. Such an element, thought of as an isometry of $Y$, cannot be parallel to any hyperplane. $\square$

[CS] P.-E. Caprace and M. Sageev, *Rank rigidity for CAT(0) cube complexes*.

[CU] M. Clay and C. Uyanik, *Simultaneous construction of hyperbolic isometries*.