There are already some excellent answers to this question, but I'd approach the answer slightly differently. Consider the circuit below.
Under normal operation (i.e. fuse not blown), Vf is IL*R, where R is the inherent fuse resistance. The current, IL, flows through both fuse and load. The voltage across the load, VL = VB - Vf, where VB >> Vf. The majority of the voltage is dropped by the load, and only a small amount is dropped by the fuse.
As pointed out by others, the power dissipated in the fuse is IL2R. At some level of dissipation, the fuse will open. As the fuse opens, an arc forms that burns away more of the fuse material. During this process, Vf will start out being IL*R (as defined above), but will become VB as IL drops to zero and fuse opens fully. At the end of the clearing event, all of VB appears across Vf and current flow stops completely.
The voltage rating (and AC/DC specification) of the fuse comes into play only after the fuse opens. A fuse with inadequate voltage rating may be unable to quench the resulting arc, leading to rapid failure of the fuse. Similarly, a fuse or breaker rated for use with AC will likely depend on a zero crossing to quench the arc, where DC-rated fuses (especially high voltage DC fuses) are often tightly packed with sand or other arc-quenching material in order to prevent the power dissipated in the arc (theoretically up to VB * IL) from catastrophically destroying the fuse and to ensure that current does not continue to flow via a continuous arc (i.e. the fuse blows yet current continues to flow via plasma between fuse internals).
If the fuse never blows, the voltage rating of the fuse doesn't matter. At the moment it does blow, the current rating ceases to matter and you'll quickly know if you spec'd the appropriate voltage fuse for your application.