Powering a 6-12V motor using a lead acid 6V battery with 6Amps of current?
If you take the "full load" to be the "stall current", then you can work out the resistance of the winding.

R=V/I = 12/14 = 0.857Ω

From that you can then work out the stall current at 6V:

I=VR = 6×0.857 = 5.143A

That, being the stall current, is the maximum the motor will draw - the current drawn when it is trying to move an impossible amount of load and the rotor cannot move.  At that current you will melt the motor drivers (or they will go into thermal shutdown).

Of course, with a geared motor, that becomes a (comparatively) massive load and would be quite hard to achieve.  You have a 50:1 gear ratio, so the stall load is effectively 50x greater.

So there is going to be a limit to the load that you can drive with the motor before it becomes a problem for the driver.  There are ways of calculating it from the given torque, gear ratio, load/noload current ratings, etc, but I don't have all that at my fingertips right now.

So you can either "play it safe" and invest in a driver that can cope with the stall current, or experiment and measure what current is drawn by the motor with different loads and see if it stays below the 1.2A limit of the shield.