Intensity of Hawking radiation for different observers relative to a black hole

This paper discusses these issues in a fairly comprehensible way. Faraway observers (like your observer A) see thermal Hawking radiation with an effective temperature given by the Hawking temperature $$T_H := \frac{\hbar c^3}{8 \pi G M k_B},$$ where $M$ is the black hole's mass.

If an observer on a string is very slowly lowered toward the black hole (so that her $dr/d\tau$ is very small), then the effective temperature increases without bound and diverges at the horizon, so your observer B inevitably gets burned up. (You can think of her as needing unboundedly large acceleration to stay out of the hole, so that she sees a huge Unruh radation. More realistically, of course, the string would break first.) See Figure 1 of the paper, which refers to the temperature observed by a strongly accelerated observer at constant $r$ as the "fiducial temperature" $T_\text{FID}$.

If an observer free-falls into the black hole, the effective temperature that she observes (which the paper calls the "free-falling at rest temperature" $T_\text{FFAR}$) gradually increases from $T_H$ very far away to $2 T_H$ at the horizon (your observer C), also plotted in Figure 1. You might think that her thermometer hitting $2 T_H$ would give her a local probe of exactly when she crosses the horizon, which would violate the equivalence principle. But this is not the case, for a subtle reason given below Fig. 1:

We note that our method gives a physically reasonable answer for $T_\text{FFAR}$ at all values of $r ≥ 2m$. However, the precise numerical value $T_\text{FFAR} = 2TH$ at the event horizon has limited operational meaning. First of all, as was discussed in the introduction, the local free-fall temperature is not a precise notion. On top of this, a freely-falling observer passing through the horizon has only a proper time of order $m$ left before running into the curvature singularity at $r = 0$, and since the characteristic wavelength of thermal radiation at $T ∼ 2T_H$ is also of order $m$, the observer cannot measure temperature to better than $O(1/m)$ precision near the horizon. Although our result for the free-fall temperature is thus only qualitative in the region near the event horizon, it does confirm the expectation expressed in early work of Unruh [13] that an infalling observer will not run into highly energetic particles at the horizon.

The case of observer C illustrates an important subtlety regarding Hawking radiation. It's often stated that Hawking radiation is just the Unruh radiation seen by an observer near the horizon accelerating away from it in order to keep from falling in. But this is not quite correct, as explained in this answer, because Unruh radiation is a flat-spacetime effect and spacetime is curved near an event horizon. If they were truly equivalent, then a freefalling observer would not observe any Hawking radiation, but in fact observer C does. But for a very large black hole, the curvature at the horizon (as measured, say, by the Kretschmann scalar $$R_{\mu \nu \rho \sigma} R^{\mu \nu \rho \sigma} \big|_{r = 2GM} = \frac{48 (G M)^2}{(2 G M)^6} = \frac{3}{4 (GM)^4}$$ ) becomes arbitrarily small. So near the horizon of a very large black hole, Hawking radiation and Unruh radiation become essentially the same thing, and indeed a freefalling observer sees negligible Hawking radiation (at an arbitrarily low temperature $2 T_H \propto 1/M$).

Edit: I should clarify that the temperature you observe as you free-fall through the horizon depends on your speed, or more precisely, how far away you were from the horizon when you were released from rest. $T_\text{FFAR}$ is the temperature that you observe if you are both inertial ("free-fall") and have proper velocity $dr/d\tau = 0$ ("at rest"). The paper https://arxiv.org/abs/1608.02532 explains this in more detail, distinguishes between the observed temperatures of the outgoing and incoming radiation, and elaborates on the distinction between the Hawking and Unruh effects. Thanks to Akoben for pointing this out.