**Edit:** The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it. 


This group is a subgroup of an [S-arithmetic lattice][1], which acts discretely on finite-valence Serre tree associated to $SL_2$ (really, a [Bruhat-Tits building][2] associated to $SL_2(F)$, where $F$ is a local field), hence is virtually free. 

The rational quaternions is a [quaternion algebra][3] with Hilbert symbol $\binom{\underline{-1,-1}}{\mathbb{Q}}$. We may tensor with $F=\mathbb{Q}(\sqrt{5})$
to get the quaternion algebra $A=\binom{\underline{-1,-1}}{F}$. The given elements $q_i$ are unit norm elements in the quaternion algebra $A$.
Since the real quaternions $\binom{\underline{-1,-1}}{\mathbb{R}}$ is ramified
(i.e. a division algebra), the algebra $\binom{\underline{-1,-1}}{F}$ is ramified at both real places (tensoring $A$ with $\mathbb{R}$ over the two embeddings of $F$ into $\mathbb{R}$. 

For all odd places (i.e. tensoring $A$ with $F_\mathcal{P}$, the $\mathcal{P}$-adic completion of $F$), the quaternion algebra $A_\mathcal{P}=A\otimes_F F_\mathcal{P}$ splits, i.e. is isomorphic to a matrix algebra $M_2(F_\mathcal{P})$. Since $2$ does not split over $F$, the algebra $A_{(2)}$ must also split, since $A$ must split at an even number of places by Hilbert's Reciprocity Law. 

The given elements lie in an order $\mathbb{Z}[\sqrt{5}][\frac12][1,i,j,k] \subset A$. For each odd prime $\mathcal{P}$, this lies in a compact subgroup of $A_\mathcal{P}$, and lies in a compact subgroup of the real places. So it must be a lattice in $A_{(2)}^1\cong SL_2(F_{(2)})$. Therefore, it acts on the tree associated to $SL_2(F_{(2)})$, described in [Serre's book *Trees*][4] Chapter II.1 (this is the Bruhat-Tits building associated to $SL_2(F_{(2)})$). 
 Thus, the group is virtually free.


The residue field  of $\mathcal{O}_{F_{(2)}}$ is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. However, it's clear that the automorphism group $A_5$ of the dodecahedron stabilizes a vertex of the Serre tree, and the twin dodecahedra should have automorphism group stabilizing adjacent vertices of the tree.


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