Random rotations in SO(3) and free group

Yes. Here's what should be a proof: the set of pairs of elements satisfying any particular relation is Zariski closed, hence has measure zero (to show that it is not $SO(3) \times SO(3)$ it suffices to know that at least one subgroup generated by two elements is free), and there are countably many relations.

See also: Epstein - Almost all subgroups of a Lie group are free (MR).