How is Q for surface acoustic wave devices measured?
The Wikipedia article has a really nice overview and drives home the most important note that there are a number of equivalent ways to think of the Q factor:


The frequency-to-bandwidth ratio of the frequency response
How under-damped the system is (how long it takes the oscillations to die down)
The rate at which energy is dissipated from a resonant system.


It also illustrates the other important point that Q factor applies equally well to any resonant system, regardless of whether it's an electrical resonance or a mechanical / acoustic resonance. In the case of a SAW resonator, of course, it's actually both.

The equivalence of the three items above is actually pretty intuitive. Let's take "Q is the frequency-to-bandwidth ratio" as the starting point. That extends to the second bullet pretty well -- an infinitely narrow bandwidth signal implies a signal that is a pure sine wave that never changes over time. If the amplitude of the sine wave changes, this means the signal has some bandwidth. The faster the sine wave changes, the wider the bandwidth. So yes, Q is both "frequency-to-bandwidth ratio" and "how quickly the oscillation can decay" because they're the same thing.

Similarly, it makes perfect sense that if you have an oscillating system that never loses energy, then it will just keep oscillating. The faster the system loses energy, the faster the oscillations will die down. So "rate at which energy is dissipated" is the same thing too.

So, whew, finally we can answer your question. I am not a domain expert in SAWs, but per the above it would make absolutely perfect sense to simply measure Q by measuring the bandwidth of the SAW resonator. As described here, that will also define how long it takes the oscillations to die down.