Here are some more details.  As John Voight said, the quaternion algebra is kind of irrelevant here.  If $\gamma$ is a regular semisimple element, then its centralizer is a torus ${\mathbf T}$ over ${\mathbb Q}$ satisfying
$${\mathbf T}({\mathbb Q}) = K^\times / {\mathbb Q}^\times,$$
where $K = {\mathbb Q}(\gamma)$ is the centralizer of $\gamma$ in the quaternion algebra.  In terms of algebraic groups, ${\mathbf T}$ fits into a short exact sequence of tori,
$$1 \rightarrow {\mathbf G}_m \rightarrow {\mathbf R}_{K/{\mathbb Q}} {\mathbf G}_m \rightarrow {\mathbf T} \rightarrow 1.$$

There aren't too many rank-one tori.  Since ${\mathbf T}$ is nonsplit, the character lattice of ${\mathbf T}$ is ${\mathbb Z}$, with the unique nontrivial action of $Gal(K/{\mathbb Q})$.  Thus ${\mathbf T}$ is also isomorphic to the norm-one torus ${\mathbf R}_{K/{\mathbb Q}}^1 {\mathbf G}_m$.

It happens that norm-one tori arising from cyclic extensions have trivial Sha.  This is in Platonov and Rapinchuk, I think, and probably goes back to Tate or something.  The basic idea is that Sha captures the failure of the Hasse principle.  In the context above,
$$Sha({\mathbf T}) \cong \frac{\left( {\mathbb Q}^\times \cap N_{K/{\mathbb Q}} {\mathbb A}_K^\times \right) }{ N_{K / {\mathbb Q}} K^\times}.$$
This is trivial, by the Hasse Norm Theorem (since $K / {\mathbb Q}$ is cyclic).

So we have 
$$Tam({\mathbf T}) = \frac{\# Pic({\mathbf T})}{\# Sha({\mathbf T})} = \# Pic({\mathbf T}).$$

Explicitly, the Picard group $Pic({\mathbf T})$ in this context is the group $H^1(Gal(K/{\mathbb Q}), {\mathbb Z})$, where the order-two Galois group acts by the nontrivial automorphism.  This is a group of order two.  Hence
$$Tam({\mathbf T}) = 2.$$