Suppose you were held by a very strong rope at constant Schwarzchild coordinate $r = 2M (1 + \epsilon)$ just above the event horizon of a Schwarzchild black hole of mass $M$. You would feel a proper acceleration of magnitude
$$a_\text{gravity}(r) = \frac{M}{r^2 \sqrt{1-\frac{2M}{r}}} \sim \frac{1}{4M \sqrt{\epsilon}}$$
from the rope (in units where $G = c = \hbar = 1$). According to eqs. (1.3) and (3.1) of this paper, you would also observe Hawking radiation with effective temperature
$$ T(r) = \frac{T_H}{\sqrt{1 – \frac{2 M}{r}}} \sim \frac{T_H}{\sqrt{\epsilon}}$$
coming out of the black hole, where $T_H := 1/(8 \pi M)$ is the Hawking temperature. The temperature and the acceleration would both diverge as you approach the horizon, but at a constant ratio
$$ \frac{a_\text{gravity}(r)}{T(r)} = \frac{8 \pi M^2}{r^2} \sim 2 \pi.$$
By the Stefan-Boltzmann law, you would observe a total emitted power per unit area (or equivalently radiation pressure, in units where $c=1$) of
$$ P = \frac{\pi^2}{60} T^4 = \frac{1}{245760 \pi^2 M^4 \epsilon^2}.$$
The total surface area of a black hole is a subtle concept, due to the spacetime curvature, so let's restrict ourselves to considering a small region just outside the horizon (i.e. a region with diameter much less than the Schwarzchild radius, which sets the curvature scale), which we can locally approximate by Minkowski spacetime.
Suppose you were to then unfurl a solar sail with mass surface density $\sigma$ (including the contribution from your own mass). The radiation pressure would accelerate you and the sail at
$$ a_\text{Hawking} = \frac{P}{\sigma} = \frac{1}{245760 \pi^2 \sigma M^4 \epsilon^2}$$
away from the hole.
We see that $a_\text{gravity}$ diverges much more slowly than $a_\text{Hawking}$ at small $\epsilon$. Indeed, if
$$\epsilon < \frac{1}{256 M^2 (15 \pi^2 \sigma)^{2/3}},$$
then the acceleration from the Hawking radiation wins, and would seem to blow you away from the hole! As a sanity check, as $M$ grows larger (cooler black hole) or $\sigma$ grows larger (denser and less efficient sail), the Hawking radiation becomes less effective at pushing you away.
Obviously this would be a ludicrous setup for a real black hole, but in principle, would it be possible to use such a solar sail to ride the Hawking radiation out and escape the hole? (Note that this idea is closely related to that of a black hole starship.) And if you did, how would a nearby free-falling observer passing you into the black hole describe the process? After all, according to the linked paper, to her the black hole would only be radiating at a quite gentle temperature of $2 T_H$ and would only provide a bounded radiation pressure.
Best Answer
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 $A_{sail} > mM\epsilon^{3/2}$.