I think Wawa's subtle message to you is to use a MOSFET instead. But, you can do this with a TIP122, but you might need to throw a heatsink on it -- depends on how much current that 12V LED strip requires. Also, I think Wawa assumed this is an RGB strip, but I don't see any mention of RGB or color, so, I will assume this is a monochrome strip.
Have a look at this schematic:
The R1 resistor value [820Ω] is for around 1A through the LED strip. You can calculate that value using the following formula:
R1 = (3.5V * 250) / I[sub]strip[/sub] where: Istrip is the current required by the LED strip.
Where did I get that voltage of 3.5? The voltage at the Base-Emitter junction [based on the datasheet] is estimated to be 1.5V [with around 1A at the Collector], so 5V - 1.5V is 3.5V
Be aware, though, because of the voltage drop from the Collector, to the Emitter [VCE(sat)], the LED strip will not glow as brightly as it would if connected directly to 12V. I can't say for sure what the expected VCE(sat) will be, because there is no mention of the amount of current this LED strip requires. But for an LED strip that takes around 1A, the typical VCE(sat) is estimated to be 0.8V to 0.6V. That means only around 12V - 0.8V or 11.2V will reach the LED strip, and that might be significant. But, on the plus side -- by not driving the LEDs as hard as they would be driven at the full 12V, might, actually, extend their lifespan 
If it is 0.8V, then, for an LED strip that requires 1A of current, that's a Power Dissipation [in the transistor] of:
P[sub]D[/sub] = 0.8V * 1A = 0.8W
So, the transistor will get a little hot. This gets worse as the amount of current goes up.
Using a MOSFET, as Wawa seems to be suggesting, will, if the correct one is selected, get rid of all of these issues.