Ohms law tells us that for a given resistance, current (and power) will increase in proportion to the voltage.
So lets put that in the context of your motor;
It's supposed to run on 24v and dissipate a maximum of 51 watts. p/v=i (p = power, v = voltage, i = current) so 51 watts divided by 24 volts equals 2.13 amps.
v=i*r ... so v/i = r (with r being resistance). So 24 volts divided by 2.13 amps = 11.29 ohms.
Assuming that the motor load is purely resistive and doesn't change (don't ask!) then at 30 volts it'll be drawing 30 volts / 11.29 ohms = 2.66 amps.
And 30 volts at 2.66 amps would be a power dissipation of 30 * 2.66 = 79.72 watts  which would be 56.31% over it's rated maximum.
But having just said all that, what the motor ACTUALLY draws (thus how huch heat it puts out  thus how hot it gets for a given amount of cooling) is also going to depend on it's load.
So in the context of your project:

you can probably throw that motor away since it's been smoked (it'll have damaged insulation and will probably fail again)

You need to decide if you want to take a chance on another one the same (but only feed it a maximum of 24v) or rethink it and use something else. From my experience with highpower RC helicopter motors it's to be expected that they'll get hot if run under load for a few minutes ... but the difference is that (a) they have fans built in to pull air through and (b) they're designed to take it (they can be far too hot to touch right after a flight). Other than that I probably can't help a lot unfortunately.