[StevenB]

Here's what I've thought through so far.
Efficiency is about power, so 90% efficiency should mean that the peak power of the motor+gearbox is 90% of the motor's power by itself. Since power is proportional to the product of speed and torque, the product of the "torque reduction" and the "speed reduction" should always be 90%.
The simplest answer is to just multiply the speed and torque by sqrt(0.9), which seems like a crude but useful approximation.

My second thought is that we can model the gearbox as simply an additional load on the motor, which causes it to have a lower free speed and lower stall torque.

If the CIM has a stall torque of 2.43 Newton-meters, and we treat the 10:1 gearbox as a load of 0.20 Newton-meters, then the stall torque of the output shaft will be: (2.43 - 0.20) * 10 = 22.3 N-m. The free speed will be 5310 RPM * (1 - 0.20/2.43) / 10 = 487 RPM. Using these numbers to calculate a motor curve gives an efficiency of 84% relative to the motor by itself.

Working this backward to get the load in terms of efficiency, the theoretical gearbox load is
stallTorque * (1 - sqrt(efficiency))

So it appears that multiplying both the stall torque and free speed by the square root of the efficiency is a pretty good estimate. I've seen quite a few resources who say to multiply speed and torque by the efficiency, so maybe I have the wrong definition?

I've seen tables which list the efficiency of various gear trains - spur gears, worm gears, etc. Can anyone tell me if those tables typically report "power efficiency" or "torque efficiency"?