There are a number of equivalent ways to think about Hawking radiation. One is pair creation, as endolith mentions, where the infalling particle has negative total energy and so reduces the mass of the black hole. Another way, perhaps more useful here, involves de Broglie wavelength. If the wavelength of a particle (not just photons, by the way) is greater than the Schwarzchild radius, then the particle cannot be thought of as localized within the black hole. There is a finite probability that it will be found outside. In other words, you can think of it as a tunneling process. In fact, you can derive the correct Hawking temperature from the correct wavelength and the uncertainty principle, without deploying the full machinery of quantum field theory.
So I guess that counts as #4 because it isn't on your list. You can't think of quantum particles coming from a specific point because you can't think of them ever having a specific location.
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
External observers and black hole formation
The event horizon is simply the delineation between the part of spacetime from which light can escape and the part of spacetime from which it cannot. In that sense, it is not directly observable, neither by external observers nor by infalling observers. Still, an external observer can observe the effects of the existence of a region from which nothing can escape.
An external observer can observe an object falling toward that region. The object's motion is increasingly slowed, and the light from that object is increasingly redshifted and increasingly reduced in intensity, until it is no longer observable for all practical purposes. The external observer never sees an object cross the event horizon, but the object quickly disappears from the external observer's senses because of the increasing redshift and decreasing intensity. This happens when the object is very near the event horizon.
That's true for any object falling toward the black hole, including the star itself — the star whose collapse forms the black hole. However, to say that the black hole never forms according to the external observer would be missing the point. The external observer sees the collapsing star quickly and smoothly disappear, again because of the rapidly increasing redshift as the "surface" of the star comes very close to the point of no return. In order for the distant external observer to continue detecting light from the star, larger and larger telescopes would need to be used in order to capture the ever-increasing wavelength and ever-decreasing intensity. When the redshifted wavelength exceeds the size of the universe, or when the intensity falls below one photon per age-of-the-universe, this clearly becomes hopeless. This occurs in a finite amount of time on the external observer's clock, so in this sense, the external observer does witness the formation of the black hole.
And remember that the event horizon delineates a region of spacetime. If we want to try to think of it as a region of space, then we need to remember that it can grow. The part of space where infalling objects become practically unobservable to the external observer at 2:00 can be larger than the part of space where infalling objects were becoming practically unobservable to the external observer at 1:00. If the external observer takes a video of objects falling toward a black hole, the video will show that the size of the crazy-region (around which the light from distant stars on the opposite side is bent in dizzying ways) is steadily growing as a result of the mass gained from the infalling objects — even though each infalling object becomes unobservable before reaching that (growing) region.
So yes, it's true that an external observer never sees an object cross the event horizon. And it's also true that an external observer does see the black hole form and grow, in the very real sense that the external observer could take a video and post it on the internet for the rest of us to watch (including seeing falling objects smoothly dwindle-and-disappear, as well as the dizzying effects on the background light from distant stars), all in a finite amount of time.
Hawking radiation
In contrast to the light emitted by the collapsing star, which is quickly redshifted to the point of unobservability, Hawking radiation persists. We can think of Hawking radiation as being emitted from just outside the event horizon (just outside the region from which nothing can escape), but unlike the light from the infalling star, Hawking radiation starts with arbitrarily short wavelengths, so that the wavelength received by the external observer is still finite despite the arbitrarily large redshift. Quantitatively, most of the Hawking-radiation wavelengths received by the external observer are comparable the size of the black hole. That's still a huge wavelength that would require incredibly sensitive instruments to detect (also because of the extremely low intensity), but it doesn't become increasingly difficult to detect (unless the black hole grows), in contrast to the light from the star which does become increasingly difficult to detect.
Altogether, a distant observer can detect the Hawking radiation even though that observer never sees any part of the star cross the (growing) event horizon. In fact, the spacetime of a collapsing star that is used to derive Hawking radiation predicts the experience of the distant observer that was described above.
Most importantly, the derivation of Hawking radiation does not rely on the perspective of any particular observer. The derivation takes all of spacetime into account, not just the part that a distant observer can see. Infalling objects cross the horizon in a finite amount of time on their own clocks, and the derivation of Hawking radiation "knows" this — just like it "knows" that distant observers never see those same infalling objects reach the horizon.
By the way, Hawking radiation can be — and originally was — derived using quantum field theory in classical curved spacetime, and that's the model assumed in this answer. This answer didn't use quantum gravity, which isn't necessary for deriving Hawking radiation and isn't necessary for this question.
Technical note about time and black hole formation
A more technical note for those who are comfortable with the concept of a spacelike hypersurface:
It is sometimes said that the emergence of an event horizon takes infinite time for a distant observer, but we need to be careful when talking about "time" in relativity. The distant observer never sees anything cross the horizon, because light cannot escape. However, there are spacelike hypersurfaces that include stuff behind the horizon and that also intersect the distant observer's worldline. In that sense, the horizon forms in finite time on the observer's clock, even though the observer can never see it. We can construct a continuous sequence of spacelike hypersurfaces (called a foliation), each one intersecting the distant observer's worldline at a particular time on that observer's clock, and each one intersecting the inside of the black hole. The black hole grows along this sequence of spacelike hypersurfaces, and this formation happens in finite time on the distant observer's clock.$^\dagger$
$^\dagger$ The details of the timeline are ambiguous, of course, because we can also construct (infinitely many!) other sequences of spacelike hypersurfaces instead. This is one of relativity's most basic lessons: "simultaneous" is generally ill-defined. We can't use a clock in one place to unambiguously assign times to events that occurred in a different place.