If we idealize things a good bit and use the simplest of physics, if you have a 2000lb 250hp car and a 3500 lb 250hp car and they both are going 90 mph at braking application point, simple physics says the brakes on the 3500 car has to dissipate more thermal energy to decrease it's kinetic energy (slow the thing down) to a given level (same for both cars...ie attaining the proper corner exit speed). For mass it is a linear relationship to go from KE pre braking to KE at exit speed. (this also is assuming equal RATE of heat dissipation from the brakes so the heavier one is going to take a longer distance to brake from V1 -> V2)
However, a point to bring up is that let's change the 2 vehicles to a 2000lb/1000hp car and a 2000lb/250hp car. The former is going to be going a lot faster (more kinetic energy) for a given stretch of straight than the 250hp car at the point of brake application and assuming equal lateral grip (ie both cars must slow to the same speed to negotiate the turn properly...same exit speed), then yes the higher hp car needs to dissipate more heat to decrement it's kinetic energy for this given turn. In this case it would be a V^2 relation ship so a little more speed at brake entry equates to a lot more energy to dissipate.
So I think a better way of addressing what 'kind' of car is harder on brakes/fluid is the kind of car that averages higher corner entry speeds for a given weight. So it's not as cut and dry but if 'more hp' means higher turn entry speeds, then sure the amount of hp will have more influence on braking due to the V^2 relationship, but high hp doesn't always mean higher speed
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But then there is the question of "There is a finite amount of thermal energy to be dissipated to brake to the proper speed". This is going to enter the braking power realm of rotor mass, caliper mass, fluid weight transfer, tires....all that good stuff that takes a ton of analysis to make conjectures about.
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