Can "premodular" be relaxed as a condition for uniqueness of Bruguieres/Mueger modularization? Suppose that C is a ribbon monoidal category with dominant ribbon functors F_1: C->D_1 and F_2: C->D_2 such that D_1 and D_2 are modular tensor categories, does it follow that D_1 and D_2 are equivalent as MTCs?  Here dominant means that every object in the target is a summand of an object in the image of the functor.
This is certainly true if C is premodular (semisimple with finitely many simple objects) as was proved by Bruguieres.  What if C is not premodular?  I haven't been able to locate a more general statement in the literature.
The particular case I have in mind is where C is the Kuperberg G_2-spider specialized to q a particular root of unity.  After semisimplification C is in fact premodular, but actually proving that is likely to be a lot of work (it would require writing down inductive formulas for simples, etc.).
 A: Noah, this is a comment to your answer to your question: unfortunately your functor
is not braided. Indeed the braiding on the square V_2 has 2 eigenvalues and the braiding on
the square of F_2(V_2) has 4 different eigenvalues..
Also, I think that TL_{-1} is related with third (or sixth) root of 1.
A: I think this is a counterexample to the result I was looking for.  Let C be the idempotent completion of TL_{-1}, the Temperley-Lieb category with loop value -1.  (Or equivalently, the category of finite dimensional representations of U_q(sl_2) where q is an third root of unity.)
Let D_1 be the quotient of C by negligibles.  This category is just Rep(Z/2) but with the dimension of the nontrivial rep being -1.
Let B be the Fibonacci category and let B' be its Galois conjugate, let x and x' be the nontrivial objects.  Let D_2 be the Deligne tensor product of B and B'.  Let F_2 be the functor sending V_2 to $x \boxtimes x'$, this exists by the universal property of Temperley-Lieb (i.e. $x \boxtimes x'$ is symmetrically self-dual and has quantum dimension -1).  This functor is dominant because $F_2(V_2 \otimes V_2)$ has every object of D_2 as a summand.
As Victor Ostrik points out F_2 is not a ribbon functor, so this is not a counterexample.
