The primary reason for diode failure is heat. Alternator diodes are pressed into a heat sink with a fan-over for a reason. The amperage rating of a given diode is based on proper thermal heat sinking. Most cutout diode conversions have little, if any, heat sink attached to the diode case.
At 15 amps the diode will need to dissipate 10 watts of thermal energy or it will overheat. (P=VA, 0.7V diode drop x 15A = 10.5 Watts) You could put a 100A rated diode in a cutout can, but if the can is only capable of dissipating 5 Watts of heat that 100A diode will fry and fail with an 8 amp load!
The original magnetic relay design of the cutout does not suffer this thermal dissipation limitation, as the points within do not present a voltage drop as a diode does.
To me, diode cutouts are just another case of newer not necessarily being better. Here's a few numbers in a scenario to illustrate why:
You're running and the battery is limiting voltage to 6.7V. You have "low draw" Halogen headlamps, 35W each, 3 amps to the ignition, 2 amps to the tail lights, and your battery is charging at a relatively low 3 amps. The total draw is 18.5 amps. (That's the smokin' hot limit for an A generator!!) The generator is producing 6.7V x 18.5A = 124 Watts of power.
NOW, you replace the stock relay cutout with a diode. @18.5 amps, the diode will need to dissipate 13 Watts. Since the battery regulates voltage in this system, the generator itself will now deliver 137 Watts to maintain the 6.7V at the battery. Your A generator will now produce that elusive, magical electric smoke. Lucas, move over, you've got company!
I'll keep my stock cutout. If you drive daylight only, or are OK with stock 21cp or 32cp lamps, and do not try to drive with a 1/2 dead battery that is drawing 15 amps, you will probably do fine with the diode modification.