Less lever travel, more grip force needed to get the same level of braking...
The amount of force of at the caliper is the amount of pressure you put at the lever, multiplied by the ratio of the areas of the caliper piston(s) and the MC piston.
So for arguments sake, lets say the stock MC piston is 1 square inch in area, and the caliper 6 square inches. The lever itself has about a 6 to 1 lever ratio.
So, you squeeze 10 lbs on the lever, the lever multiplies that to 60 lbs.
The MC to caliper multiply by 60 again, so you have 360 lbs of force on the brake pads.
The movement of the lever is reduced the same way the force is increased, so the end of your lever moves 36 times further than the piston moves.
If you use a 1/2 square inch area MC piston, then the force at the caliper is 720 pounds.
If you use a 2 square inch MC piston, then the force at the caliper is 180 lbs.
So, the area of the original piston is Pi/4 x D^2 = 122.7 mm^2
The new MC is 154 mm^2
So it will take approx 25% more lever pressure to get the same stopping force.
The lever will move 75% of the distance it did before.
So if the end of the lever moved an inch before, now it will only move 3/4 of an inch
If it took 20 lbs to lock the brakes, now it will take 25lbs.
Basically, it will be harder to stop. Less effective braking because you need more hand pressure.
Brakes that only needed 2 fingers to stop would require more fingers...LOL
I would try to stay the same....I think 11mm is available, and that would give you more braking power.