When you look at a spherical body, you don't "see" a sphere, you see a disc.
The same would be true of a neutron star. The difference is that the angular radius of the disc is larger than $R/D$, where $r=R$ is the coordinate radius of the neutron star and $D$ is the distance to the observer.
The "effective" radius is given by $R (1 - R_s/R)^{-1/2}$, where $R_s$ is the Schwarzschild radius for a neutron star of mass $M$.
Here is an example of what a neutron star might look like (by Corbin Zahn, from this website). This is a neutron star with radius twice the Schwarzschild radius (for example, with a radius of 8.4 km, for a $1.4 M_{\odot}$ neutron star). Each patch is a 30 degree by 30 degree square and you can see how both poles are clearly well inside the visible disc.
Should the radius shrink below 1.76 times the Schwarzschild radius (and there are equations of state for a neutron star that would permit this), then the whole of the surface would be visible (e.g.Pechenick, Ftaclas & Cohen 1983).
