Yes, I have used this as an example to illustrate that it is useful to distinguish things which allow to make fast, efficient, accurate computations from things which allow to improve conceptual understanding or to prove consistency of the theory.The question is if Gribov copies are a problem or not.
If you think that gauge-equivalent gauge fields are really identical states, and your gauge fixing condition is purely technical, then Gribov copies are clearly a problem, they mean that the same state is counted several times.
If you, instead, consider gauge-equivalent fields as physically different states (even if you have no way to distinguish them by observation), and the gauge condition as a physical equation for these degrees of freedom (so that you use a nice looking equation like the Lorenz condition), then there is no problem at all with Gribov copies.Note that the EM field is a vector field, not chiral. So, to implement an exact gauge symmetry on the lattice is not a problem.
The only question would be if this is compatible with the approach to fermion doubling, where I have the electroweak doublet instead of a single Dirac fermion. In
http://arxiv.org/abs/0908.0591 there the idea comes from I have an exact gauge invariance for U(3), but with the condition that all parts of an electroweak doublet have the same charge. While the group is fine (the SU(3) plus U(1)_{em} part is in reality U(3) too) its representation is not. But the EM field can be understood as a deformation of the diagonal U(1) symmetry. And for such a deformation the exact gauge symmetry remains, even if somehow deformed, so one can hope that the deformed symmetry remains an exact symmetry. Then, the field would remain massless.
But this is, of course, a point which needs better understanding.