The question was somewhat sloppy so it was unclear if it was about some or all von Dyck groups admitting embeddings in $U(n)$. If the question was some then Victor's answer clearly suffices.
If the question was about all von Dyck groups, then Victor's answer almost gets there but not quite, since only finitely many of these groups are arithmetic (first observed by Takeuchi who actually listed them all, but also follows from far more powerful finiteness theorem of Borel and Prasad: Only finitely many arithmetic groups with covolume $\le const$). For non-arithmetic groups one has to worry about all Galois conjugations yielding noncompact forms of $SO(3)$. For non-arithmetic groups, one has to use a bit more elaborate version of Victor's answer. First, observe that every von Dyck group $\Lambda$ contains a closed surface subgroup $\Gamma$ of finite index. I will consider only the case when the genus is $\ge 2$ since virtually abelian case is much easier. Then, being a closed surface group, $\Gamma$ is isomorphic to a cocompact arithmetic subgroup $\Gamma'$ of $O(2,1)$. Now, $\Gamma'$ admits an embedding $\rho$ to some $U(n)$ as explained by Victor. To embed (unitarily) the original group $\Lambda$, use the representation induced from $\rho: \Gamma\to U(n)$ to $\Lambda$.
A more interesting question is if every finitely generated subgroup of $SL(n, {\mathbb R})$ embeds in some orthogonal group. It seems that $SL(m, {\mathbb Z})$ is a counter-example for that.