To drive that LED to it's specified capacity you'd want to push 6A into it. If you are willing to sacrifice some (maybe most) of it's very long lifetime, you can push it beyond that, probably well beyond, but I haven't seen much info on overdriving that particular device. It may be out there, but I just haven't seen it.
On the other hand, at that power level, a host that small is not going to keep it cool very long. If you want to push it to even 6A, you'll need thermal step-down either in the driver or the operator (when the light gets hot, turn the brightness down).
As to the load on the battery, in your case a reasonable assumption makes it really easy. Since the LED voltage is around 3V, and the battery voltage is higher than that, but not by much, you can assume that whatever current you send to the LED is the same current you need to draw from the battery. We could go through a lot of complicated explanations and calculations, but at the end of the day we'd end up right about there.
So if you want to drive the LED at 6A, you need a battery that can source 6A. Nowadays, most quality 18650 cells can do that no problem.
But that would be considered a moderately high drain, and you should consider what that drain does to the capacity of the cell. For any given cell, the higher the load you put on it the lower the capacity is. "High drain" cells are designed to minimize this effect. The VPCA5 is a very high drain cell, and loading it at 6A, you will probably get its rated capacity or better. A cell rated at 8 or 10A loaded at 6A would likely give you well below its rated capacity. So if you paid a premium for a 3500 mA-H cell, but got one rated at 8A, you might not get much or even any better performance than you'd get from the VPCA5, rated at 2600 mA-H and capable of 35A.
I'm sure that doesn't sound as clear to others as is does in my head, so please ask for clarification as necessary.