What's the problem with Black Hole evaporation? Black hole evaporation is not unitary because it takes a pure state to a mixed state. On the other hand, ordinary decay processes in Quantum Mechanics do not seem very unitary either. (For example, if I know my final state is a Breit-Wigner distribution, I don't think it's possible in general to reconstruct the initial state.) In what way are decays any more unitary than BH evaporation?
 A: 
On the other hand, ordinary decay processes in Quantum Mechanics do not seem very unitary either.

No, decay processes are perfectly unitary. However, they entangle the decaying object and the environment, so you can't recover the state by measuring the object or environment alone. 
As a simple example, consider an atom with ground and excited states $|g\rangle$ and $|e \rangle$, inside a cavity which can have $|0 \rangle$ or $|1 \rangle$ photons in it, at frequency $\omega = (E_e - E_g)/\hbar$. The part of the Hamiltonian responsible for decay is 
$$H_{\text{int}} = \alpha |1g \rangle \langle 0e | + \text{h.c.}$$
and the evolution is perfectly unitary. For example, starting from the state $|e0 \rangle$ one smoothly evolves into a superposition of $|e0 \rangle$ and $|g1 \rangle$. The description of other quantum decay processes is similar, except that the environment will have more degrees of freedom. The state is pure at all times.

In what way are decays any more unitary than BH evaporation?

In the standard semiclassical derivation of Hawking radiation (which does not look at all like finding $H_{\text{int}}$ above) one finds that the emission spectrum is exactly thermal. So when the black hole completely evaporates, you're left with only thermal degrees of freedom, which correspond to a mixed state. That's the puzzle.
A: In QFT with classical gravity, the problem with Black Hole evaporation is that the asymptotic spectrum is that of a thermal density matrix with respect to the free theory of an asymptotic observer. The asymptotic density matrix does not merely have Planck-spectrum expectation values - it is also completely diagonal, meaning that the coherence between the off-diagonal matrix elements vanishes.
This is because the particles escaping to infinity are entangled with the particles falling in to the black hole, and since there is no way to measure what fell in to the black hole, the reduced density matrix on the Hilbert space of particles escaping to infinity is that of a mixed state.
On the other hand, in a decay, the evolution is unitary for any finite time. The state will be the Breit-Wigner, plus some exponentially decaying transients. At any finite time, the exponentially decaying deviation from Breit-Wigner can (in principal) be measured exactly to reconstruct the original state.
