Equatorial radius (km) 6,378
6,378 + 300 km plug in Surface of a sphere = 4Pi * r squared equals 560,405,420 Sq km
6,378 + 600 km equals 611,887,283 sq km
So there is 51,481,863 more sq km at the higher orbit or there is about
10% more area at higher orbit.
What determines whether something hits the Hubble is it's target size (in a 3d space, this roughly translates into volume) over time compared to number of potential impactors crossing the Hubble's path.
I'm not going to do the math but basically if there are the same amount of objects in the higher orbit which has 10% more area it should decrease the chance by more just 10%.
There are other factors, such as a lower orbit you will have less natural impactors because the earth and it's atmosphere will block them or manmade objects will remain in a higher for longer period of time
since they have a lower decay rate, and highly elipical orbits would have less chance of impacting because they have lower percentage of time crossing the Hubble's path.