The absorption and emission of photons during bound-bound transitions in atoms is perfectly well described by single-photon physics, with no soft photons involved.
The fancy QFT mathematics in Jeanbaptiste's answer go beyond my expertise, but they deal with bremsstrahlung-like processes with unbound electrons, and it is missing the wrangling of QED required to describe bound states. In any case, QFT is not required to describe atomic transitions unless you are doing spectroscopy at high levels of precision – and, even then, you're still calculating small corrections to the energy of the (single) photon involved.
More particularly, the specific concerns that you posed do not justify your conclusion that "in practice some energy is always lost in the form of low-energy photons", which is not itself the same as "transferred into heat".
Im more detail:
- there is always energy mismatch between a photon and an atom (e.g., due to the atom thermal motion)
There can be an energy mismatch between the energy of the transition in the laboratory frame and the Doppler-shifted energy in the atom's rest frame, and this Doppler shift is perfectly easy to account for, as it is purely kinematic.
There is also a nontrivial dynamical effect in that the absorption or emission of a photon delivers a nonzero kick to the atom's centre of mass. This can be completely accounted for within standard atomic physics (I explained the details in this Q&A), and the result is simply a shift of the transition energy. In other words, the transition remains a single-photon process, and the only effect is a change in the energy of the photon.
- atom is coupled to the vacuum photon modes, which results in broadening of the transition
- the absorption happens over finite time
These two statements are simply Fourier transforms of each other. The atomic bound "eigenstates" are only eigenstates of the atom-only Hamiltonian, but they are not eigenstates of the full Hamiltonian of the system. (Otherwise, they would not decay!) Instead, once you account for coupling to the electromagnetic fields, they become resonances, with a finite width and a finite lifetime. But the coupling is still a single-photon one, and no soft photons are required to explain this.