The only thing that I can think of is to apply a high enough current load to the output so that the voltage drop across the resistor is significantly higher the Vf of the diode. That will allow you to read the diode Vf and delay time. OTOH I don't know if that will affect other devices in your circuit. In addition, I would pulse the diode circuit at a very low duty cycle ( 0.1 to 1.0%) to minimize heating of the circuit components. I would start with a 10 Amp current so 10A x .2 Ohm = 2 volts at the input and short the diode/resister back to the return side of the pulse generator. 10 amps would create 2 volts across the resister alone and the diode should clamp at more or less .6 volts so it would be easy to see on a scope. Dealing with that low of a circuit resistance would mean that you would need a pulse generator with a very low output impedance and VERY good conductors connecting it to the circuit being tested. Even a 0.1 Ohm lead resistance would throw your results off by 50%. What you really need is a kelvin connection (4 wire) from the pulse generator to the circuit under test to compensate for the lead losses, but I've never seen that on a pulse generator.
FWIW I've never dealt with high speed relays but you might be able to use a standard kelvin connected power supply and control the power to the circuit via switching relays. A lot is going to depend on how robust the other devices in the circuit are and how much you need to limit the total energy into the circuit.