Electronic – Most efficient/quietest way to drop DC

A zener diode is absolutely no good for very low power (for the 1N4732A the zener voltage is specified at 53mA), and even LDOs often have ground currents 10 times your 5\$\mu\$A load. You want an LDO with a < 1\$\mu\$A ground current, like Seiko S-812C40. Output voltage is 2.0 to 6.0 V, selectable in 0.1 V steps, so there's also a 4V type. You get 65mA out. Dropout as low as 120mV, and stable even without output cap. The S-812C is available in SOT-23.

edit
I just discovered that there's also a version with a shutdown. That S-812C is even cooler than I thought! :-)

If by "best" you mean "most efficient", then you want a switched-mode power supply, or more specifically, a buck converter. Efficiencies above 70% are common, and above 90% is possible with careful design and selection.

With such a device, you may indeed be able to use a 12V battery without any more loss than you would have with a 6V battery.

The LM317 is a pass transistor driven by an error amplifier, essentially an op-amp plus a transistor. An op-amp plus a MOSFET is not very different, and you will not improve on the efficiency of an LM317 with any combination of op-amps and MOSFETS that don't involve rapid switching, and probably an inductor. Without switching, you have a linear regulator, and the only place excess voltage can go is into heat.

Electrical power is the product of current and voltage:

$$ P = IE $$

and as long as you have any device with a voltage across it and current flowing through it, you are converting electrical energy into something else, usually heat. Non-linear regulators avoid this problem by rapidly switching between periods of high voltage and low current, and low voltage and high current, except in an energy storage device (an inductor) which can store the electrical energy in something more easily recovered than heat (a magnetic field).