How do I terminate the isolation port to prevent reflections? Various sources online just say that I should terminate with the characteristic impedance
And that would be correct, any output with a characteristic impedance can only be terminated properly by terminating it with a resistor to ground with the same value as the characteristic impedance, so 50 ohms in this case.
The ground should be the same ground as the shield of the signal output. Use any other ground and the signal path will be longer which adds inductance to the 50 ohms which will prevent a proper termination.
What if you do not properly terminate the signal? Then the output signal at that port will reflect back into that same port. Terminate it properly and all signal will be converted to heat by the resistor. Then (almost) no signal will reflect back.
but is it sufficient to place a short microstrip at the isolated port and then via to a common ground plane?
A microstrip is just a path for the signal to a different location. It is not a proper termination. At the end of the (50 ohm) microstrip you would still need a 50 ohm resistor.
Is it bad to use the common ground plane?
It depends on your design, if you are making a sensitive (easily disturbed) RF circuit then you don't want to share the same ground (plane) with some digital circuits. That would increase noise levels. If your RF signals are quite large it might not matter.
Do I need to worry about RF vias here, or is a regular via sufficient?
My guess is that an RF via has less inductance. Less inductance is almost always a good thing in RF design, so if you have the option: yes, use RF vias.
I've seen a design that uses a 0402 49.9ohm resistor between the isolated port and a via to ground (connected by short microstrips). Does that provide any additional protection?
Protection for / against what? This is the proper way to terminate a 50 ohm (49.9 ohms is close enough to 50 ohms) port. Just do the same!