The TIP120 does not need 120mA at the base for normal operation, that's the absolute maximum rating, above which you don't want to go.
The spec you are mostly interested in is the hFE (current gain), which for a darlington is very high, since it's two transistors connected in a way so the current gains multiply.
For the TIP120 it's given as minimum 1,000 (compare with a typical 200 for a single bipolar transistor)
Also important are the max collector current (5A) and the collector emitter voltage (60V)
The main disadvantages are that the base-emitter voltage is doubled compared to a single transistor (~1.4V), and the saturation voltage is higher (typically ~0.8V compared to ~0.2V at low currents)
These points are rarely a problem for a simple switch driven from a micro pin. At higher collector-emitter currents though, the Vsat rises and can interfere with desired operation and cause problems with dissipation.
For example, in the TIP120 datasheet note that at 3A Ice, Vsat is given as 2V, but at 5A it has risen to 4V. That's 20W of dissipation, a lot to heat to try and get rid of to keep the temperature down. So when switching a large current you need to take these factors into account, and maybe decide to look at a more suitable part (e.g. logic level, low Rsdon power MOSFET)
Since we have a gain of 1000, we hardly have to draw anything from the micro pin. Let's say we want to switch 1 Amp:
1A / 1000 = 1mA into the base needed.
If we have a drive voltage of 5V, then we subtract the Vbe from the drive voltage and divide by the current:
(5V - 1.4V) / 1mA = 3.6k resistor. To give it a bit of leeway select something a bit smaller like 2.2k. This still only draws ~1.6mA.
I wouldn't read too much into the different prices - the price of components is often dictated by how popular they are, the more they sell the less they cost. If you see better specs at a cheaper price, go for it ;-)
You can some pretty odd prices when the component is scarce/new/obsolete - I saw a 10uF ceramic capacitor priced at £7.50 (qty 1) on Farnell the other week...
You might need a bigger transistor. For large motors google "darlington" to get more current gain. Also scope the output of your arduino to see if it really works. No scope: add a second 1kohm resistor feeding a capacitor (few uF) to the output of the arduino and measure with a DMM. At 100% dutycycle you should get close to supply voltage of the chip, at 50% half, etc. Check chip hardware and/or code to make sure you don't have inverted logic: ex 80% duty cycle is not 80% LOW and 20% HI as if driving a PNP transistor.
Best Answer
The TIP120 will work at 3.3V as well. You do not apply a voltage to a bipolar transistors base, but you apply a current to the base. The resistor between your micro controller and the TIP120 will convert the voltage to a current. You might need to reduce the resistance your base resistor a bit at 3.3V.
Using a Darlington transistor can be effective for loads with a high voltage that are switched often, but are only switch on for a short time. If you switch a low voltage load, e.g.: a single LED, a motor running at 3.3 V volts, then bipolar transistors and especially darlingtons will be very ineffective. MOSFET can often be just as easy (or even simpler) to use and losses at the transistor will be smaller. Between collector and emitter of a Darlington transistor you will always lose around 1.2 V, no matter how much current you are switching. This is especially problematic if you are operating at high currents and low voltages. See for example these tutorials on how to use nFETs as a switch: http://www.hobbytronics.co.uk/arduino-tutorial9-power http://bildr.org/2012/03/rfp30n06le-arduino/