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 $\le const$). In this case, 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 far more interesting question is if every finitely generated subgroup of 
$SL(n, {\mathbb R})$ embeds in some orthogonal group. Even the case of $SL(3, {\mathbb Z})$ is unclear as there are no "obvious" orthogonal representations (with infinite image). The positive answer might follow from Lubotzky's characterization of finitely generated linear groups.