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.
Hopefully I have understood your situation correctly, if not then please let me know and i'll delete this answer.
For a radially in-falling observer to hover at $r = 2M + \epsilon$, they would need to provide an opposing acceleration of $a \sim \frac{1}{4M\sqrt{\epsilon}}$, as you stated.
Any amount more than this critical amount will cause the observer to move radially outwards, away from the black hole. If the observer had a rocket strapped to her back, then she would indeed be propelled outwards by the additional acceleration provided by the hawking radiation, and any infalling observer would see nothing special: just a radially boosted observer sailing past.
The rope in your problem makes this tricky however. If the rope is tied to some fixed point far from the black hole, then you can not ride the hawking radiation from the black hole. This is because the moment you move radially outwards from your position, the rope will stop providing any force, as it presumably goes slack. Thus, gravity will immediately grab a hold of you again and drag you back to $r = 2M + \epsilon$.
If your rope is tied to an accelerating rocket or you happen to have luckily grabbed on to the tentacle of a giant galactic space squid desperately trying to save you, then you will again be in the situation where it is as if you have a rocket attached to your back and an in-falling observer will again see nothing weird.
With regards to whether or not you can escape a black hole eventually powered only by Hawking radiation, you might want to take a look at the calculation that has been done in this paper (see also this paper).
Conceptually, a hovering observer will measure hawking radiation and this will give an acceleration to the observer, potentially allowing them to escape the black hole. However, with regards to turning off the acceleration that keeps you static, you would need to wait until the amount of hawking radiation was large enough to sustain your motion without the tension of the rope.
This may in principle be possible, but you would have to wait for an incredibly long time, since black holes evaporate slowly and give out very little hawking radiation until they have very small mass. At this point, quantum gravity becomes important so who knows.
Another point is about the sail. By calculating the acceleration at $r = 2M(1+\epsilon)$, you are only getting information about what the acceleration would be there at some fixed $\sigma$, which would die out as you got further away. In order to calculate this more effectively. To maintain the same acceleration, you would need to increase the area of your sail as you moved out.
The power is given by
$$
P \propto A_{BH}T(r)^4 \propto \frac{1}{M^2(1-\frac{2M}{r})^2}
$$
While the acceleration by
$$
a_{hawking} = \frac{P}{\sigma} = \frac{A_{sail}}{mM^2(1-\frac{2M}{r})^2}
$$
Where $\sigma = m/A_{sail}$.
This needs to be bigger than the gravitational acceleration felt by the observer, meaning our area needs to be:
$$
A_{sail} > \frac{mM^3(1-\frac{2M}{r})^{3/2}}{r^2}
$$
For $r = 2M(1+\epsilon)$, this means that $A_{sail} > mM\epsilon^{3/2}$.
Best Answer
So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.
Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.
The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.
There's another reason why your logic is failing.
The entire idea of Hawking radiation depends on the existence of a curved space time and an event horizon. But when $M=0$ we just get a flat spacetime. There is no event horizon. And note that an $R=0$ event horizon would mean there was nothing inside the black hole - no volume, nothing.
The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $M\to 0$. Again, you're using an approximation based on the assumption of a curved spacetime ($M>0$) and applying it outside it's "designed purpose" as an approximation.