Why is the chiral symmetry only $SU(3) \times SU(3)$ and not $SU(6)$?
Okay, I think I have an answer but I would interested to hear some feedback on this. I initially thought that by writing the left and right handed fields in 2 component notation ($q_L $ and $q_R^c$), they were 'equivalent' objects and can freely rotated between one another. However, I realized now this is not the case.
The left and right handed fields are in the fundamental and anti-fundamental representation of $SU(3)_C$ (due to the often omitted conjugate symbol on the right handed fields) and so can't be simply rotated between one another. This restricts left fields to rotate into left fields and right fields and right fields and hence the $SU(3) _L\times SU(3)_R $ instead of a rotation between all the fields ($SU(6)$).
Note that for the above discussion I was using common beyond Standard Model notation where the fundamental fields are all left-chiral fields and hence transform the same way (for details on this see hep-ph/9709356, bottom of pg 8). An alternative to this viewpoint is to define the fundamental fields as left and right chiral fields and then as @Thomas mentioned they transform the same way under $SU(3)_C$ but you can't rotate between them since they have different representations under the Lorentz group.
On second thought, the reason the Noether currents and charges of all such continuous transformations vanish, as per my comments to your answer, @JeffDrorr, is that the transformation $\delta \psi_L = \theta \psi_R$ simply fails to exist, at all, and is made to appear to exist by the admitted lack of intuitive transparency of Dirac's notation!
Consider a free Lagrangian of a left-handed fermion, $ i\overline{\psi_L}~ \partial\!\!/ ~~\psi_L = i\overline{\psi_L}~P_R\partial\!\!/ ~ P_L\psi_L $, where the r.h.side is written in perverse redundancy, at no extra cost, by the idempotence of chiral projectors! Now, apply the L to R transformation the scheme is predicated on to the last fermion, to generate an "increment" $i\overline{\psi_L}~P_R\partial\!\!/ ~P_L \theta ~ \psi_R $, stillborn by the action of the projection operator. Well, given the left-hand projector preceding it, you never really transformed the field: the O(θ) term is not there, before we even go onto transforming the $\bar\psi$s, let alone a complement of a free right-handed fermion lagrangian whose right handed fermions would rotate to left handed ones. This is not really a symmetry under a transformation, it is invariance under stasis!
The internal symmetries combined with this fundamental non-transformation are tensor-multiplied red herrings. I confess such are the hardest problems to resolve, namely studying unicorns which do not exist, because they couldn't....