BCS theory is able to derive the London equations and the Meißner-Ochsenfeld effect. Experimentally the Meißner-Ochsenfeld effect can be demonstrated via levitating superconducting rings. So we have the usual Lorentz force acting on the Cooper pairs carrying the current. However in order to lift the ring the force has to act on the ring i.e. on the lattice as a whole. But there is no interaction between Cooper pairs and the lattice due to the energy gap seperating the BCS ground state from the 1st excited state. So my question is how the force can act on the ring w/o having any interaction between the Cooper pairs and the lattice? Let me speculate that due to the disperison relation of accustic phonons ##\omega(k) = c_S\,k## there is a coupling of two effective d.o.f., namely the ground state of the Cooper pairs and the lattice, which allows for a collective mode at k=0 w/o energy gap for the phonons. This would correspond to a movement of the lattice as a whole. EDIT: any ideas regarding this critical paper? http://iopscience.iop.org/1402-4896/85/3/035704/
I don't know the precise answer to your question. However, it is well known that the simple BCS treatment isn't precise enough to describe completely the interaction of the cooper pairs and the magnetic field. There are Goldstone boson like excitations which become massive due to the Anderson-Higgs mechanism. Nambu has first given a description which is gauge invariant and also contains these long range modes.
I just came back from vacations and found time to look up some reference which I found quite lucid: Cremer, S., M. Sapir, and D. Lurié. "Collective modes, coupling constants and dynamical-symmetry rearrangement in superconductivity." Il Nuovo Cimento B Series 11 6.2 (1971): 179-205. It is an easier read than Nambu's original article. I found it quite interesting as it also showed how to treat bound states in QFT, something which is missing in most books on QFT. The most interesting point in the whole treatment is the following: Why aren't there any Goldstone bosons in the reduced BCS hamiltonian? Because the reduced hamiltonian corresponds to a nonlocal long-range interaction which allows to circumvent the Goldstone theorem! When treating the full Hamiltonian, there are indeed long range interactions, but the reason that there are also no Goldstone bosons is now the Higgs mechanism!
I thought a little bit more about your question and I think the collective modes calculated e.g. by Nambu are not the ones relevant as they are also gapped. In fact, a careful examination of even the BCS hamiltonian shows that is has some ungapped modes which the following guys call the "thin spectrum": van Wezel, Jasper, and Jeroen van den Brink. "Spontaneous symmetry breaking and decoherence in superconductors." Physical Review B 77.6 (2008): 064523.