If I use a diode I will lose line voltage ~0.6-0.7V, thats cutting it really close for operational voltage.
I realize there is some current feedback when using a resistor but 10mA or less is very small, and the USB ports are almost certainly already feedback protected.
No not at all, it will not burn off the difference, you should get more familiar with ohms law.
ex. one is 5.1V and the other is 4.9V, we have one resistor value of lets say 20ohms isolating them. That's a difference of 0.2V across the resistor, with 0.2V/20ohm = 10mA current flow, which puts us at 2mW power dissipation on the resistor.
But please tell me, what is wrong with this ohms law math? Do you know how voltage and resistance works in a circuit?