First: If you have a motor rated at 12VDC, then it's never a good idea to heavily decrease supply voltage. Even not if the load is minimal. The reason is, that every motor needs a minimum torque to begin the motion. And with 3V this is just too little.
Do always connect to your motor the nominal voltage, that the motor is designed for. And if the speed is too high, then use a suited gear, as Steve already mentioned. But do not heavily decrease supply voltage.

For a 'naked' DC motor there are mainly two ways to regulate speed against load changes. If motor current is very high and heat dissipation of electronics would waste lots of power, then a discontinuous regulation can be used, where the motor is connected to supply voltage via a switch for short periods of time. Regulation is done by varying the duty cycle of this switching.
A suited regulation circuit 'looks' at the voltage generated by the motor, when it is disconnected from supply voltage, means when the motor works like a generator. As then no relevant current through the motor is flowing, observable voltage across motor terminals is directly proportional to the speed!

Unfortunately such a design needs a good and ripple free lowpass filter, a PI-stage, a sawtooth generator and two switches: One that connects and disconnects the supply voltage to and from the motor terminals and one that disconnects and connects the regulator input from and to the motor terminals.
Of course, all the signal conditioning can be handled by a 8052 microcontroller..

If heat dissipation of electronics does not play a significant role, a continuos regulation can be used. Here the motor is permanently connected to its supply. Not directly, but via a small series resistance. If under full regulation once a stable condition has reached, then every change of load (load current!) will result in a change of voltage drop across this series resistor, and by this in a change of motor voltage, which can then be used to keep the speed constant.

A very simple but rather good working circuit is the follwing:



Whenever the load current increases, voltage at motor decreases. This is sensored by emitter of BC107, which potential also decreases. By this Ube of BC107 increases and turns on MJE2955 a bit more. As consequence voltage at collector of MJE2955 increases a bit to compensate for the increased voltage drop across 2R7 resistor.

Maybe such a simple speed control is sufficient for your application?

Kai