Ask Electronics

Fire hazard? Not really. The charger will not provide more than 5 V, 3.1 A or whatever its rating is. Even if the strip fails short circuit (unlikely, most LEDs blow open and there is at least a resistor in series anyway), nothing will receive more than 15 W of power. A 1m cable can safely dissipate that.

Worst case scenario, an LED fails short and its series resistor will receive over 6x its intended power (with 5 V across it instead of 2 V, and its current will increase correspondingly). It is soldered on power-dissipating flexible PCB and will probably not blow open. It will get the area somewhat hot and potentially melt the plastic in the back of the monitor while the rest of the strip keeps glowing but more dimly. Hard to tell if it could get over flammable temperatures with moderate heatsinking and only a few watts of power.

To keep safe, I would deliberately add resistance before the LED strip, using something like a 1Ω 5W resistor (or a shitty long cable). That way, the voltage drops significantly in case of a short. Also, the LEDs will run at lower-than-intended current, which prolongs their life and decreases risk of failure.

Edit: Some microwaves have a 0.8Ω 25W resistor as part of inrush, at least in 230V regions. Feel free to use that, it will happily handle a semi-short circuit. Or you know, an automotive fuse.

The safest option is replacing the whole setup with an LED strip that has no resistors (bare LEDs) and a constant current driver.