On the 'idle' graph I make it 12% duty cycle. So since current is 0.5A peak, that makes it:
0.5*0.12=60mA average
0.5*0.12^0.5=173mA RMS.

Can't really find anything useful from the other graph - it just shows how at that speed the current is limited by the inductance of the motor. I guess we could calculate the motor inductance ... T=2.07ms, change in current is 700mA, so dI/dt=700/2.07=1.449A/s (linear approximation).. applied voltage is 24V minus a bit for losses so call it 23V, so the inductance L=24/1.449=15.9mH (roughly, oversimplifying a bit there). That's either a pathetic motor or I've done something wrong! Unless it's wired in bipolar series, that would be reasonable since the same motor in bipolar parallel would be about 3mH which is not bad.

Either way these numbers seem a bit fishy.. please could you link to the motors you are using (datasheet?) and confirm that the resistor you're measuring across is definately 0.4ohms, not two 0.4 ohm resistors in parallel or something like that.

The simple answer is just increase the current and so long as the motors don't get too hot, i.e. above 80°C on the case, it's fine.