Your question is related to Schwarzschild's spacetimes, which are two spherically symmetric solutions of the equations of General Relativity.
The first one is the exterior Schwarzschild solution (eSS) which describes the curvature (gravity) produced by a spherical object out of it. It is a solution of the vacuum, so that it does not explain the gravitational field inside the sphere. The latter could be modelled with the interior Schwarzschild solution (iSS), if we assume that the sun is a sphere of a perfect fluid with a radial distribution of density and pressure.
Now, let us consider the two important distances in the situation: the radius of the initial star, $r=R$, and the critical radius for the collapse, $r=R_{Sch}$, the so-called Schwarzschild radius that corresponds to the horizon of the black hole (for the real sun it is around 3 km!).
Obviously, $R_{Sch} < R$ (in other case we would have had a black hole from the beginning), so the universe can be separated into three regions:
I ($0 \leq r < R_{Sch}$),
II ($R_{Sch} \leq r < R$) and
III ($r \geq R$).
And you will have the following distribution:
- Before the collapse: iSS (I, II) and eSS (III).
- After the collapse: eSS (I , II, III). There is vacuum everywhere except at the singularity, and now the region I becomes exotic (it is the no-return zone of the black hole).
(Answer) As you can see, in the exterior of the initial star (the region III) the spacetime remains exactly the same, as well as the orbits of the planets.
Note. Both solutions (iSS and eSS) can be "pasted" together at the point $r=R$ in a continuous and regular way.