Measure the resistance of the motor. (Do this a few times and rotate the shaft, and take the lowest value, because the brushes can interfere with getting a good measurement).
Dividing 18V(or your actual supply voltage) by this resistance gives you the starting current; your motor driver needs to withstand that.
Say you measure 2 ohms; the starting current would be 9 Amps.
Two ways of reducing this current : run the motor off a lower voltage, or use PWM.
A series resistor is a traditional way of starting a motor, but it should be switched out of circuit as soon as possible once the motor is running. PWM during start is better and more efficient.
EDIT : given Spehro's revelation on the controller IC, the right answer might be to keep it and add external MOSFETs (it is after all a MOSFET driver!) capable of handling the current you need.
EDIT for SherrellBC's questions...
1) I am not going to recommend a way to disconnect a series resistor once started. It is a traditional "big motor" (think tramcar!) approach but there are better approaches here.
2) True the inductance limits current momentarily then the current settles to 9A. However this is NOT the steady state current - unless the motor is stalled. And if stalled - unless the motor is specially designed for such abuse - you had better disconnect the current quickly!
Then the rotor starts to rotate. And this is what you are missing : when it rotates it also acts as a generator, and the voltage it generates (called "back EMF) opposes the driving voltage. This reduces the current.
If the motor is lightly loaded (as it should be for efficiency) it will spin up to generate (say) 15V, leaving only 3V across its 3 ohm resistance, and consume 1 amp.
Load it a bit more, and it will draw more current (say 3 amps) to generate the extra torque. It slows down so that the back EMF is reduced to allow the extra current draw.