Is the radiation problem actually solved in the classical quantum model of hydrogen? It's often said that in classical physics a electron-proton system is not stable due to "Bremsstrahlung" and that one instead has to look at it quantum mechanically.
This doesn't make sense to me. The quantum mechanical Hamiltonian doesn't account for "Bremsstrahlung" either. Is this taken care of only in QED?
 A: The stability of the hydrogen ground state follows from the Schrödinger equation predicts a lowest bound state, the ground state. A treatment including radiation does not change the fact that there is no lower state to which the system can progress. Of course , the question now is why the Schrödinger equation is correct. To this there appears to be no answer at present.
A: 
It's often said that in classical physics a electron-proton system is not stable due to "Bremsstrahlung" and that one instead has to look at it quantum mechanically.
This doesn't make sense to me. The quantum mechanical Hamiltonian doesn't account for "Bremsstrahlung" either. Is this taken care of only in QED?

The reason for the first fact is people know that Bohr gave a convincing argument for why the classical Coulomb-force model can't be correct (because it ignores radiation and its destabilizing effect), and erroneously think that later quantum theory does not have or have solved that problem.
Of course, the standard model of atom in non-relativistic theory does not address or solve the problem at all. There is no radiation in this universally accepted model at all! Bohr's/Schroedinger's model of atom is stable for the same reason non-relativistic Coulomb-force atom model is stable or Newtonian solar system model is stable: no force retardation, no radiation, no relativity is allowed.
When we acknowledge relativistic aspects of EM interaction, the question of which states are "stable"  becomes more involved. Now the Hamiltonian isn't so simple and it isn't even clear we have the right Hamiltonian.
To my knowledge, there is no complete proof of stability of hydrogen atom in QFT allowing all relativistic aspects of EM interaction. Quantum field theory of bound states is hard and most published work makes additional assumptions. There is the Bethe-Salpeter equation which is proclaimed as general, but assumptions are always made to get solutions. Those are: positronium isn't stable but hydrogen atom is (technically "resonance" vs "bound state"). Details such as particle being proton instead of positron play big role here. Is muon-electron system stable? Proton-muon system? Experiments tell us likely answer and we bend the theory to accomodate. There isn't first principles based, dynamical reason for why some are stable and some not.
One often reads about there not being a lower energy state than the ground state as the ultimate reason. But this rests on assumptions about how the complete Hamiltonian looks like. In particular, quadratic Hamiltonian in field strength is almost universally assumed. As is well known, this Hamiltonian poses various problems with infinities. What is less universally known is that this quadratic Hamiltonian is an additional assumption on top of Maxwell's equations and relativity theory and does not derive from them. There may very well be a different way to analyze consequences of EM theory and relativity than using that Hamiltonian.
A: The quantum mechanical model of a bound electron-proton system does not include brehmsstrahlung because the electrons are not small balls orbiting the nucleus.  They exist in stationary energy eigenstates, and do not emit any radiation unless they are making a transition.
At the most simplified level, we could simply couple the classical electromagnetic field to the quantum mechanical model of the atom by inserting the expected values of the charge and current densities into Maxwell's equations.  If one does this, then one finds that the Larmor radiation formula yields a radiated power which depends on $\frac{d}{dt}\langle\mathbf p\rangle$.  For an energy eigenstate, $\langle \mathbf p \rangle=0$, so no radiation is generated.
As a more sophisticated model, one could "second-quantize" the electromagnetic field and couple the single-electron Hilbert space to the photon Fock space.  In this picture, if we restrict our attention to the space of states in which the electron is in an "old" energy eigenstate, then the zero-photon state of the electromagnetic field is an effective ground state, and no brehmsstrahlung photons are emitted.
That being said, such a state is not a true eigenstate of the full Hamiltonian if the electron is not in its ground state, and vacuum fluctuations can induce transitions in which the electron moves to a lower energy state and the photon number increases by one - this is spontaneous emission.
