We can represent the collapse of an object (e.g. a star) by a Penrose causal diagram looking like:
The trouble is that I now have to explain what a Penrose diagram is, but that would be a long article in itself. The vertical axis is a timelike coordinate and horizontal distance away from the vertical dashed line is sort of radial distance. One of the key points about these diagrams is that light rays always travel at 45º. You'll see why this matters in a moment.
The green line shows the surface of the collapsing object, and the red line shows the singularity. You might be a bit puzzled that the singularity is a line not as point, but remember that this is a spacetime diagram so every point in space traces out a line as it moves through time.
If we consider a light ray emitted from the surface of the star, then that light ray will travel at 45º from lower left to upper right, like the blue arrow I've drawn. It should be obvious that every light ray emitted after the blue arrow must hit the singularity, so the blue arrow shows the time at which the event horizon forms. It should also also be obvious that this happens before the singularity forms.
So the answer to your question is that yes the horizon does form before the singularity forms.
The vertical axis is not time as you and I understand it. It's actually the timelike coordinate $v$ from the Kruskal-Szekeres coordinates that has been further transformed to map the infinite time axis onto a finite line. The mapping preserves causality, so something at later "time" cannot cause something at an earlier "time". Thus we can confidently state that the singularity forms "after" the horizon. But this does not correspond in any useful way to times you might measure with your clock. In fact according your clock neither the horizon nor the singularity will ever form i.e. both would require an infinite time to form.