This group is a subgroup of an [S-arithmetic lattice][1], which acts discretely on finite-valence Bass-Serre tree, hence is virtually free. 

The unit quaternions $q_i$ may be interpreted as elements in $SU(2)$, treating left multiplication as unitary transformations of $\mathbb{C}^2$. The entries to the matrices lie in $\mathbb{Q}(\sqrt{5},i)$, and traces lie in $\mathbb{Q}(\sqrt{5})$. 
Then $tr(q_i)=-\frac12$, and one may apply Theorem 3.6.2 of [MacLachlan-Reid][2]
to conclude that the [quaternion algebra][3] generated by this group over $\mathbb{Q}(\sqrt{5})$ has Hilbert symbol $(-15, -110)$. Thus, the quaternion algebra ramifies at both real places. I consulted Matt Stover at lunch, who also explained to me that it is ramified at both places sitting over the prime $11$, and no where else. So the group acts discretely on the Bass-Serre tree associated to the prime $2$ (which doesn't split in $\mathbb{Q}(\sqrt{5})$). Thus, the group is virtually free.

I'm not sure how to prove that it's freely generated by $q_i$, without explicitly computing the action on the Bass-Serre tree. One might be able to take advantage of the $S_4$ symmetry of the generators to determine this. I think the Bass-Serre tree will be defined by orders in the quaternion algebra over $\mathbb{Z}[(1+\sqrt{5})/2][\frac12]$ (or maybe more properly working over the 2-adic completion). The residue field is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Bass-Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Bass-Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. 








  [1]: http://en.wikipedia.org/wiki/Lattice_(discrete_subgroup)#S-arithmetic_lattices
  [2]: http://books.google.com/books?id=yrmT56mpw3kC&lpg=PP1&pg=PP1#v=onepage&q&f=false
  [3]: http://en.wikipedia.org/wiki/Quaternion_algebra