Component leads and PCB trace inductance are many times greater than the squiggle OP is chasing after in a 2010 resistor.
Just get an RF resistor usable to GHz instead of designing and measuring parasitics to save pennies.
Would you think this is realistic, based on your power experience?
Is there such a thing as an "RF" resistor that's not in the 50-100 ohm range?
The underlying pattern is this: the resistor is simply some conductor over the ground plane. The higher above it is, and the narrower it is, the higher the transmission line impedance is.
There isn't such a thing as an "RF" 1mΩ resistor, because no one has a transmission line that matches 1mΩ. (There can be a resistor with Kelvin leads, with matched mutual inductance, such that the current-sense bandwidth is flat into the GHz; but the inductance of the main current path, will necessarily be limited by physical dimensions, period.)
We're talking short transmission line segments here (<< 1ns), but the conclusion is the same: make it wider and lower, and you'll be better off than you were. In short: use "wide" resistors (say, 1020 vs. 2010), use more in parallel, and use more vias flanking the common pad (or even via-in-pad if your soldering process or fab cost can support it).
Incidentally, a lot of "wide" resistors are trimmed in the same way as usual (i.e., with slots that neck down the current path, increasing inductance). Perhaps try the Susumu(?) resistors, that claim to be many segments in parallel, in a wide body?
Beyond that -- if this is, like, GaN switching shit, there's not much you can do as long as the components have to remain on the top surface of the PCB. Even if you integrate resistance into the PCB itself, you still have the stray inductance of pads, traces and vias, carrying that current from the device, into the depth of the PCB. There is only one possible solution at this point: choose a lower switching current, with more channels acting in parallel. This relaxes the impedance demands per channel, making the layout tractable.
Tim