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 $\mathrm{d}r/\mathrm{d}\tau$:
$$\left|v_s\right| = \sqrt{\frac{2M}{r}}\text{.}$$
