Let's say you assume that the neutron star is spherically symmetric, e.g., ignore the effects of rotation. Then for a radial trajectory in the resulting Schwarzschild spacetime, the calculation is actually not quite wrong, although you must be careful in interpreting it.
The reason is that orbits in a Schwarzschild spacetime have an effective potential that conserves an analogue of specific orbital energy:
$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$
where $l$ is specific angular momentum. The difference with the Newtonian case is that $r$ is the Schwarzschild radial coordinate, not the radial distance, and $\tau$ is the proper time of the orbiting particle, not absolute time, and finally, the very last term is absent in Newtonian gravity.
For a radial orbit, $l = 0$. Therefore, if by $v$ you mean the rate of change in Schwarzschild radial coordinate with respect to proper time of the particle, then the Newtonian calculation is actually correct:
$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2M}{r}}\text{.}$$
However, if $v$ means something else, such as the rate of change in Schwarzschild radial coordinate with respect to Schwarzschild time $t$, then
$$\left|\frac{\mathrm{d}r}{\mathrm{d}t}\right| = \left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}\text{.}$$
Finally, the velocity measured by a stationary observer at a given Schwarzschild radial coordinate can be calculated via an inner product between the stationary observer's four-velocity $\left(1/\sqrt{1-2M/r},0\right)$ and the particle's four-velocity $\left(1/(1-2M/r),\pm\sqrt{\frac{2M}{r}}\right)$, but in this case it happily enough turns out to be equal to $\mathrm{d}r/\mathrm{d}\tau$:
$$\left|v_s\right| = \sqrt{\frac{2M}{r}}\text{.}$$
Your argument is true classically also. You see, the effect of Gravitation close to earth and that of acceleration in flat spacetime is same. But this equivalence goes way past just the elevator experience, where all the observer feeling is a state of weightlessness or equal weight. While introducing GR Einstein took this equivalence farther and said that there are no laws of physics that can distinguish between an accelerated frame and a stationary frame in gravitational field. This goes for all laws including Electro-magnetism. So yes, if you see a charged particle in accelerated frame it should radiate. This can then be extrapolated and we can say that a charged particle which is stationary in a gravitational field should also radiate.
Best Answer
I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates):
$$ ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1} $$
with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass:
$$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)c^2dt^2 + \frac{dr^2}{\left(1-\frac{2GM}{c^2r}\right)} + r^2 d\Omega^2 \tag{2} $$
The difference is that factor of $1-2GM/c^2r$, which we can also write as $1-r_s/r$ where $r_s$ is the Schwarzschild radius - $r_s = 2GM/c^2$. Feeding in the mass and radius of the neutron star we find this factor is about $0.72$, so general relativistic effects are indeed important.
Your question (1) is answered in What is the weight equation through general relativity?. The coordinate acceleration measured by an observer at the surface is:
$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} \tag{3} $$
so it differs from the Newtonian prediction by (in this case) a factor of about $\sqrt{0.72}$.
Re your questions (2) and (3), offhand I don't know the expression for the coordinate acceleration measured far from the star, but it will not be the same as equation (3). A distant observer sees falling objects slow as they approach the event horizon and asymptotically approach zero speed at the horizon. So the coordinate acceleration is obviously different from the coordinate acceleration measured near the horizon.