Coulomb's Law modified in general relativity? It seems difficult to track down a clear explanation of this statement:

So although the Coulomb law was discovered in a supporting frame, general relativity tells us that the field of such a charge is not precisely $1 / r^2$.

Papers I've found seem to say either that the inverse square law for the Coulomb field of a massive point charge remains exactly true in general relativity, or that it has corrections on the order of $1/r^4$ and higher. 
I suspect that I may be getting lost in the coordinate transformations and the in-context meaning of $r$.  What I would like to know is if general relativity predicts any deviations from the inverse square law for the electric field surrounding a point massive charged particle, as seen by a distant observer. 
Presumably the measurement method should be specified, so here's a possibility: Attach a charge $Q$ to an extremely massive particle, then probe the electric field of the charge by first measuring the distribution of electrons, then of positrons, shot at various energies toward the massive particle a la the Rutherford experiment. The difference between the two would be used to subtract out the purely gravitational attraction between the massive particle and the probe particles. I realize that this approach would be impractical for measuring extremely small deviations from Coulomb's law, but it should at least provide a way to dodge some of the difficulties associated with defining $r$ near a point mass.  I also realize that quantum corrections would totally change things in real experiment.  I'm just looking for a clear classical explanation of what happens to Coulomb's law due to the effects of gravity per general relativity.
EDIT 12/24/18: Specifically: in the scattering measurement proposed above, does it appear to a distant observer that there is a deviation from the inverse square law - a deviation that is a bit different from that due strictly to gravity?
 A: The Reissner-Nordstrom metric describes a massive point charge in General Relativity. The covariant component of the electrostatic four-potential, $A_t$, is exactly $Q/r$, but the contravariant component $A^t$ is more complicated. Similarly, the purely covariant component $F_{rt}$ of the electrostatic field, and also the purely contravariant component $F^{rt}$, is exactly $Q/r^2$ but the mixed components are more complicated.
This is in Schwarzschild-style coordinates where the area of a sphere around the charge is $4\pi r^2$ as $r\rightarrow\infty$.
There is no meaningful way to say which components are the potential and the field. The potential is a four-vector and the field is a four tensor, and they have both covariant and contravariant components which are equally valid.
Regardless of which components you use, the electric flux through a sphere at infinity is $4\pi Q$.
When you think about Coulomb’s Law, you should not think of it in the “high-school” form that the field is inverse-square. Instead you should think of it in the “university” form that $\nabla\cdot\mathbf{E}=4\pi\rho$. This equation, generalized to take curved spacetime into account, remains true when the effect of gravity is considered.
A: All of these answers miss the essential point.
Coulomb's law relates to electrostatics: what is the force between two charges at rest? Equivalently, what is the electric field of a point charge at rest? For the sake of this question (and said wikipedia page), we neglect the mass of the point charge: the R-N metric does NOT apply. Instead, we are trying to calculate the field in a "supported" frame in a weak (by GR standards) uniform gravitational field.
Why is such a frame of interest? Because the surface of the Earth (where Coulomb's law was discovered) is such a frame. In Newtonian gravity, the surface of the Earth is an inertial frame (neglecting Earth's rotation and motion through space); objects on the surface experience a gravitational force and a balancing "normal force" of non-gravitational origin. But recall that there is no such thing as a "gravitational force" in GR. We should be free-falling towards the center of the Earth; we are not because we (and everything on Earth's surface) are "supported" against this free-fall by other forces. Therefore our frame is in uniform acceleration (relative to the local spacetime); the applicable metric is the Rindler metric.
Back to the Coulomb field: an observer in flat space measures a perfect inverse square field. Does a supported observer (as defined above) measure that as well? The answer is no: there are slight deviations from the perfect inverse square law - even though both the observer and the charge being observed are equally supported. By dimensional analysis, the first-order correction is of order [g ⋅ r / c2], with g measuring the strength of the local gravitational field (i.e. the g00 component of the metric), r the distance from the charge where the electric field is measured, and c the speed of light. So the field in the supported frame would be E ~ q (1/r2 + O[g ⋅ r/c2]). This effect has nothing to do with the question of measuring the 'r' coordinate in a non-flat spacetime.
