Fuses blow based on the current flowing through them.
The heat produced in a resistive element is I^2R...the higher the current, the greater the heating.
Normally the load will limit the current flow, and the heating in the fuse is essentially constant for the same current flow. A 1 A fuse in a 12V circuit with 750mA flowing through it (9W dissipated in the load) will have essentially the same heating as a 1A fuse in a 240V circuit with 750mA flowing through it (180W dissipated in the load). In the event of an overload (say 1.5A flowing), the fuses will experience the same heating.
In the higher voltage system, more power is being delivered to the load, both during normal and overload conditions. The fuse will reach its trip point in the same way at both voltages.
I would expect a fuse to _open_ differently with different supply voltages. As the fuse opens, more and more of the supply voltage is dropping across the fuse, and less is being dropped across the load. As this happens, the heating in the fuse increases, and the supply voltage will determine the ultimate heating that is possible before the current flow actually stops. I expect that the details of the different operating characteristics are quite complex: high voltage would mean that the metal melts faster, but could mean that arcs get sustained for longer periods.
In any case, when a fuse is rated at a given voltage, it may be used at lower voltage, all other things being equal.
Of greater concern is the difference between DC and AC; a fuse that works just fine with 240V AC supply might not be able to function safely with a 60V DC supply. This is because once the metal melts you must also deal with quenching any arc that forms.
-Jon
