Is no-load to stall amp draw linear with DC motors?
In the steady-state case, it's a simple linear interpolation:

\$I_S\$ is the stall current.

\$\omega_{nl}\$ is the no-load speed.

\$I_{nl}\$ is the no-load current.

For a motor loaded such that it spins with a speed \$\omega_L\$, the current draw is:

$$I_L = (I_S- I_{nl}) (1-\frac{\omega_L}{\omega_{nl}}) + I_{nl} $$

For example, you have a motor with a 2.2A stall current, 0.2A no-load current, and let's assume the no-load speed is 1000 rpm. Technically \$\omega\$ usually refers to rads/s, but here the units cancel out; the only important factor is what the relative speed of the motor compared to the no-load speed is.

At a 250 rpm load speed, the current draw is:

$$I_L = (2.2A - 0.2A) (1-\frac{250}{1000}) + 0.2A = 1.7A$$

Keep in mind that even though the steady-state current of the motor is 1.7A, the peak current could be much more. Starting up the motor would draw close to the stall current of 2.2A, decreasing until the motor reaches steady state. Depending on how quickly the motor is able to get to steady state, this may be significant or not.