I guess I should have made my conclusions explicit:
Classically, the escape velocity is independent of the direction of emission, because the gravitational force is potential (unlike, say, magnetism or friction). In GR the situation is more complicated because of the potential capture by an event horizon.
Light always escapes, regardless of direction (assuming your cylinder is transparent), if there is no horizon close by. In other words, the only time a ray of light can be captured is when it dips under the event horizon. This is basically the definition of the event horizon.
The anisotropic light ray capture happens when light is emitted close to the horizon, such as some light rays are bent hard enough to go through the horizon.
Quite independently of all this, any attempt to create a cylinder such as you describe will fail because it will collapse onto itself before you can make it heavy enough.
Classically, the escape velocity is independent of the direction of emission, because the gravitational force is potential (unlike, say, magnetism or friction). In GR the situation is more complicated because of the potential capture by an event horizon.
In a real world, the escape speed from the system (Earth)-(any black hole) is heavily dependent from the direction you choose to escape. In the direction from Earth to the black hole, it is greater then c. While in the opposite direction it is only the well known 11+ km per second.
What bothers me further...
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?