Extracting back EMF from DC motor
Is this true? If it is so, is there anyway this could be done?


No.

It actually boils down to simple basic physics.  There are a tremendous number of power conversion problems that can be solved simply by applying conservation of energy.  This is one.

Imagine you have a perfect, lossless motor.  No resistance in the wires, no hysteresis or eddy currents in the armatures, no mechanical friction or other losses.  If you ignore energy stored in inductances or spinning mass (or if you carefully take them into account), then electrical power in + mechanical power in = 0.  (I'm writing it this way to make it equally useful for generators).  Write that as $$P_{e} + P_m = 0$$

Mechanically, power is equal to \$P_m = 2\pi T_a \omega\$, where \$T_a\$ is the torque exerted on the rotor (or armature, for an old-style brushed DC motor), and \$\omega\$ is the speed of the shaft.

Electrically, power is equal to \$P_e = i_a v_a\$, where \$i_a\$ and \$v_a\$ are the current and voltage in the armature.

Back-EMF implies that there's a non-zero \$\omega\$.  If there were some means where you could bring the back-EMF to zero with a non-zero torque and speed, then you would have \$P_e = i_a \cdot 0 = 0\$.  But then you would have \$T_a \omega + 0 = 0\$ for a non-zero torque and speed -- you'd be getting energy out of nothing (or expending energy into nothing).

The un-handy part of this is that humans still need to generate energy.  The handy part is that the equations that I've just laid on you apply to all electric motors -- DC brushed, DC brushless, universal wound, AC induction machines, etc.  If there's electricity going in, and mechanical energy coming out, then to the degree that the motor is efficient, you can predict its behavior -- and you can generally come pretty close by assuming a 100% efficient motor and then fudging for the inefficiencies.