A few things:
1)Just because an observer crossing the event horizon doesn't necessarily feel ill effects AT THE TIME OF CROSSING the horizon, it doesn't mean that they won't inevitably end up at the singularity, where there will be plenty of ill effects--all timelike curves that cross the horizon end up at the singularity in a finite amount of proper time. For a particle falling into a non-spinning black hole, it's actually the same amount of proper time that it would take to fall into a Newtonian point mass.
2) You have to be very careful about what you mean by 'horizon' in the case of a black hole that eventually evaporates. There are several definitions of 'horizon', and depending on how you resolve the singularity, and upon how the hole evaporates these different definitions can differ in meaning--the most common difference is the apparent horizon-a 'point at which, for this given time, you can't go back', and the event horizon--'the point at which, you MUST end up at the singularity'. It might be possible that your evaporating black hole spacetime may have an apparent horizon but no event horizon, for instance. In that case, the whole paradox goes away.
3) A careful answer of this requires the careful drawing of a Penrose-Carter diagram of the relevant spacetime. If you managed to tweak it somehow so that you fell in, blasted your rockets for long enough to outlive the recontraction of the horizon, the short answer is that you wouldn't receive all of the information about all of the future, just that determined by the "null past" of the horizon--you would find out about all of the lightlike and timelike rays that fell into the horizon, but not those that would head toward the spot where the horizon used to be at times later than when the horizon was there.
This is not a full answer, since I don't know the full answer, but it is more than a comment.
My contribution is to compare your question with another, simpler one, and then come back to yours.
1. Observer-dependence of radiation in classical physics
Here is the simpler question (one to which an answer is known). A charged particle accelerating in empty space will emit electromagnetic radiation. A charged particle fixed at some point in a static gravitational field will not emit electromagnetic radiation. Both statements are true---relative to a certain natural choice of reference frame in each case. But, in the first example one could step on board a rocket accelerating with the particle, and no electromagnetic waves would be seen in the rocket frame. In the second example, one could go freely falling past the particle, and in this freely falling frame, electromagnetic radiation will be seen. So what is going on here? Does the charged particle emit radiation or doesn't it?
All of this scenario can be treated with special relativity, and it teaches good lessons to prepare us for general relativity. The main lesson here is that the process of emission and subsequent absorption of radiation is not absolute but relative, when accelerating reference frames are considered, but the state changes associated with absorbing radiation, such as a detector clicking, are absolute. It is the way we interpret what caused a detector to click that can change from one frame to another.
To make the above really connect with your question, note that in my simple scenarios I could imagine a cloud of gas possibly absorbing the radiation in between emitter and receiver, just like in your scenario.
2. To resolve a paradox in observer-dependent physics, first convince yourself about the easiest observer, then seek arguments to explain what the other observer finds
The above principle can be applied to resolve puzzles in relativity such as whether a fast pole can fit into a short barn, or whether a fast rivet can squash a bug in a hole.
3. Unruh effect has two complementary physical interpretations, depending on who is accelerating
Hawking radiation is like Unruh radiation, and therefore it is more subtle than classical radiation. A useful tip from the consideration of Unruh radiation is as follows. Unruh's calculation says a detector accelerating through the vacuum picks up internal energy, equivalent to detecting particles. Now if we look at this detector from an inertial frame, we still conclude it picks up excitation, but we interpret differently: we say the force pushing it provided some energy which got converted into internal energy because the process is not perfectly smooth.
4. Answer
Now I will apply all the above to provide an answer to your question.
I admit I am not sure and the following answer is just my guess. I only claim it to be an intelligent guess.
My guess is that the distant observer observes Hawking radiation and absorption lines. I say this because it is a consistent summary of what seems to me to be ordinary physics, assuming that there is Hawking radiation coming up from a horizon.
So the puzzle is to explain this from the point of view of the freely falling cloud. I think an observer falling with the cloud looks up at his distant friend and notices that his friend is accelerating through the vacuum, and consequently experiencing internal excitation owing to fluctuation of the forces accelerating him. To account for the absorption lines, i.e. the absence of excitation at certain frequencies, I guess (and this is the speculative part) that now the calculation from quantum field theory would have to take into account that the rest of spacetime is not empty but has the cloud you mentioned, and this cloud affects the overall action of the quantum fields in this way. Note, it is not a case of action at a distance (in either perspective), but it is a highly surprising prediction so I think your question is a very interesting one.
Best Answer
Yes, the atom will be torn apart, and eventually in a Black Hole (BH) get ripped apart radially (to the BH) and get compressed into nothingness perpendicularly.
Even before that happens the gravitational tidal force will rip off the electRons and have the nucleus break up and have most of it converted to neutrons, and then rip those off and get to the quarks, and eventually fall into the singularity. That's why we say that a BH forms when there is too much gravity, nothing can withstand the gravitational effects. Not electron pressure (which holds up white dwarf stars), not nuclear forces (neutron stars), and not strong forces (quark stars, or some parts of the cores of neutron stars).
Yes, the equivalence principle says everything will be accelerated (i.e. pulled) the same way, but only until the force differential between two objects in the atom see different accelerations - that's the gravitational tidal effects, caused by very strong curvatures of the spacetime due to gravity.
Now, that all according to classical GR. As they get closer to the singularity in the BH quantum gravity effect will increase, and we don't know what eventually happens because we still don't have a theory of quantum gravity. But the atomic components all get obliterated as explained above. Quantum gravity effects enter in after that, as the gravitational effects become even stronger, and smaller distance scales get affected. The characteristic distances at which this happens is on the order of the Planck length, about $10^{-33}$ cms, much less than the nuclear or quark scales.