Help with TIP122 darlington transistor
What you seem to have built is an emitter follower, where the emitter follows the base voltage, minus the base-emitter voltage. This configuration is also known as common collector. This \$V_{BE}\$ will be around 1.4V since the Darlington gives you two PN junctions between base and emitter. The reason the emitter voltage doesn't get higher: if it would there wouldn't be enough voltage drop across the base-emitter junction to cause any base current, and hence no collector current either. That's why the emitter voltage is stuck at about 4V.
Like Leon suggested, place the motor on the collector side, and connect the emitter to ground, so that you get a common emitter configuration. You also need a base resistor. The datasheet says that \$V_{BE}\$ can be as high as 2.5V, so


  \$ R = \dfrac{5V - 2.5V}{3mA} = 833\Omega \$


That's a maximum value, I would pick a 680\$\Omega\$; the 1.6k\$\Omega\$ is too high to give you 3mA of base current. Don't forget to place a diode in parallel with the motor, anode to the transistor's collector.  

edit
Something important I forgot to mention: at 3A the transistor may have a \$V_{CE(SAT)}\$ of 2V, which gives you a 6W power dissipation. That's a lot and requires a sizable heatsink.
Russell also calculates the dissipation, but arrives at other values because he works with typical values, whereas I calculated for maximum values. You always should calculate for worst case. A heatsink you picked for 3W dissipation will be insufficient in case the dissipation is 6W, which may happen. That's why it is mentioned as maximum value.  

A better alternative for the Darlington is a MOSFET with a low \$R_{DS(ON)}\$, like the FDC653N. Thanks to its low \$R_{DS(ON)}\$ of 0.055\$\Omega\$ it doesn't need a heatsink if it's mounted on sufficiently large copper. You may find MOSFETs with even better specs.  



Further reading
More on common collector
More on common emitter