I'm provisionally convinced that, for a cylinder of nonzero thickness, you can indeed find some combination of density, length, and thickness such that the escape velocity is greater than c longitudinally from the endcap, but not perpendicularly from the midpoint.
I find this hard to believe, actually. Granted, I have not done the calculation, but in my mind the gravitational potential profile is such that the ends are always at a higher potential than the middle, so the classical escape velocity should be less from the ends. Unless the cylinder is of non-uniform density, which would be a totally different problem.
Oh, and classically the escape velocity, which is just sqrt(2*potential), is independent of the initial direction, provided the trajectory does not hit the ground.
For a cylinder of zero thickness, you are correct. The trick is to have nonzero thickness. This doesn't (to first order) affect the gravitational attraction at the endpoints, but it limits how close you can get to the axis. Even if you allow a particle to enter the cylinder, it will then no longer feel the attraction from the outer parts. It therefore seems to me that you can choose a density and radius such that there is never an event horizon as you go perpendicularly in, but a length such that there is one at the endpoints.
I have no answer for the argu...
As mister shminux mentioned somewhere, he is happy and qualified to answer questions in the field of the Relativity. Here is mine:
A long rod (a cylinder) could have a large escape velocity in the direction of its main axe. From its end, to the "infinity". Larger than the speed of light. While the perpendicular escape velocity is lesser than the speed of light.
Is this rod then an asymmetric black hole?