If I have a Gaussian beam and I use a spiral phase plate (SPP) to convert it into a vortex beam, I get a donut-shaped beam with zero intensity at the center. Imagine that I design an obstacle which fits exactly the intensity profile so that all the light on the bright ring gets totally absorbed, but the dark area is completely unblocked. On the output I must get "nothing", darkness. Now, my question is: what if I let this "nothing" go through SPP back-converter? Will I get light back out of "nothing"?
The SPP is passive optics: it (ideally) moves light but neither absorbs it nor creates it. The light from the centre is moved towards the external ring (which is wider than the original Gaussian beam). If you further add an absorbing obstacle down the beam, there is no way a SPP recreates the light.
However, your description of such an obstacle is wrong: to have “norhing” at the output, you need to have 100% absorption everywhere (except on a single point, at the centre, but that is not physical). Obviously, if you have such a totally absorbing obstacle upstream, you can’t have anything downstream which recreates the original beam.
However, you may want to approximate the above absorbing profile with something which absorbs everything in the ring, except in a small window around the centre, to only keep the “doughnut hole”. The intensity in this hole is weak, but non-zero and is typically $∝r²$, where $r$ is the centre of the beam. This leads to a total intensity $∝R⁴$ in a window of radius $R$. This (weak) beam is still essentially a vortex beam, with the same phase profile, so, the SPP back-converter will convert it back to an approximately Gaussian beam, with no hole. But this beam will be very weak, having no more light than the one that passes through the obstacle, since the SPP does not creates “light out of ‘nothing’.”