Dude.. you are going to use 2*AA li-ions on that flashlight, right?
Thats 8.4v charged, 7.2v nominal and 6.4v discharged. The diode only needs 2.5v so, you will be turning extra voltage into heat. That isn't good as 5.9v + current would be turned into heat.. I don't know if the heat generated would be the current draw of the diode, if thats the case, the average heat generate @ 300mA output, would be 1.7W... nasty.
I recommend you should use a single 4.2 li ion, a spacer (to fill out the empty area) and a resistor to limit current. Even without the resistor your diode should work.. for a very little lifetime, thats why you should use a resistor.
IF you want a driver.. You should get a flexdrive and TWO common AA.. IMO...
edit: answer to your second post:
Dude, the Rckstr need 1.5v extra voltage so
2.5v (diode voltage) + 1.5 = 4v. So as two AA gives out 3v they won't work. A single 14500 will work like a charm (charged) and two will generate MUCH heat.
oh btw.. please write everything in a single post. Don't post many things like I do