Electronic – the simplest and cheapest way to convert 11.1VDC battery pack to regulated 9VDC

The simpliest schematics is probably amplifier from any ion smoke detector. Connect base of n-p-n darlington (the one without internal resistors) to metal box and emitter to ground. Apply 50% of transistor rated c-e voltage (10..20v) through 1..10k resistor with LED.

For fun, try 4-6 transistors in enourmous darlington chain with LED. It will work like "detector of everything". In my experiment it detected a lit match few feet away with "antenna" 1 inch long. (Sorry no proof of working in vacuum due to budget limitations)

In terms of efficiency, this is about as good as it will get. The most efficient method of transforming the voltage from 12V to 5V is to use a buck converter as opposed to a linear regulator. The UBEC is a buck converter. For power levels in this range, an efficiency of over 90% is considered good.

If you were willing to search around for a custom buck controller IC and do some PCB design, you could potentially eke out up to 96% by selecting an IC that includes a technology called synchronous rectification. An example of such an IC is https://www.fairchildsemi.com/ds/FA/FAN2001.pdf (however note that the input voltage on this IC only goes up to 5.5V)

With synchronous rectification, you can still achieve a good efficiency at low current levels (see page 5 in the pdf). For 100mA you can still get well over 90% efficiency.