Here's an idea.  It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis.  Now, here are two things which I think are true but which I don't know how to prove:

- The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is trivial.
- $\text{SO}(3)$ is equal to its own commutator subgroup.

If both of these things are true, it follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i \in G_{n-1}$ but $w \not \in G_n$ for some $n$ and setting the $w_i$ to be rotations about a common axis.