no, it is not true. the following is already contained in Andreas Thom question.
from the first paragraph of his question:
Let $n$ be an integer. Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$.(In fact, he showed that any finite subgroup of $GL(n)$ has a abelian normal subgroup of finite index, bounded independently of the subgroup.)
Take $u_1,u_2\in{U}(n)$ that generate a free group (easy to construct for $n\geq{2}$), and let $m$ be as above. Then, since $v_1^m,v_2^m$ commute, $$ \|u_1^mu_2^m-u_2^mu_1^m\|\leq{}2\|u_1^m-v_1^m\|+2\|u_2^m-v_2^m\|\leq{}4m\varepsilon$$
Since $\epsilon$ is arbitrary we have a contradiction.
