First of all, in general you will get more light output with an LED having a higher luminous flux bin. Thus, a bin number closer to the end of the alphabet would produce more light at a given current level, but these LEDs are rare and highly sought after. You will also have more difficulty finding the color bin you desire in a higher output LED because everyone wants them!
"Better" might depend on where the Vf (last of the four digits in Luxeon codes) for your particular TXOK falls. It also depends on the current level at which you would like to operate the LED.
If you intend to use a 3.6V rechargeable, and if the Vf for your particular "K" voltage binned LED is higher, then the circuit will kick up the voltage to control to 750mA (some QIIIs run at a different current). In this mode the circuit will convert about 17% of the energy passing through it into heat (83% efficiency, which is generous for the stock QIII circuit). So, in the case that Vf for your "K" bin is 3.75V, then the LED consumes 2.81W at 750mA and the circuit dissipates about an additional 0.58W for a total of 3.39W. If the Vf for the LED is lower than 3.6V, then the circuit should go into bypass mode and direct drive off of the battery voltage. This may result in overdriving your LED, which is not recommended with the stock QIII heatsink (I made a custom copper one instead).
I chose to purchase a "J" bin for Vf and ensure that the circuit would not be needed (I removed it too to make room for more copper) by direct driving with a resistor to set the current. For example, assuming 3.6V from the battery and a drive current of 750mA and you have a Vf at the lower end of the "J" bin = 3.27V, then you will need a resistor of 0.44 Ohms which will dissipate 0.25W as heat. (Use LED Pro 2.12, courtesy of Jtice, which can be found on this site to simplify these calculations) In this case the LED running at 750mA consumes 2.45W for a total of 2.70W being drawn from the battery. Thus, direct drive can conserve battery power compared to situations where the regulation circuit is needed. Less energy would need to be dissipated through a resistor for Vf closer to V battery, but for an LED with a higher Vf a higher voltage would be required to achieve a certain current, meaning that the LED has a higher "resistance" (an LED is non-ohmic though) and so more heat will be dissipated through the LED as I squared times R.
So, you also need to consider that the amount of heat that will be generated by the LED is influenced by the Vf bin. The rate at which the battery is drained can also be affected by Vf bin, depending upon whether you use a higher Vf with a regulating circuit or not. Since in direct drive the current at the LED is equal to the current through the resistor, the total heat generated by the circuit is not affected by choice in Vf in this case, but where would you rather have that extra heat? If you need a hint consult the light output vs junction temperature curves from Lumileds. This is why overdriving should not be done without good heat sinking: you might run the LED at a higher power level but not create any additional light while shortening it's life!