In a D.C. motor, is the Lorentz force on the conductor directly proportional to the back E.M.F.?

In a D.C. motor, when there is a current in the conductor a magnetic force arises on the conductor. If there is no load on the D.C. motor the magnetic force will move the conductor. With the motion of the conductor a changing magnetic field will create a back E.M.F. in the conductor in the direction that would cause a current in the opposite direction of the original current.

My question, then, is without a load on the D.C. motor would the magnetic force be directly proportional to the back E.M.F.?

Furthermore, if a load was introduced, The magnetic force would remain the same, given no change in current. But as a result of the more slowly changing magnetic field the back E.M.F. would decrease in magnitude. Given the smaller back E.M.F. and therefore smaller energy loss, since the original current has less opposing potential, Does this mean that a D.C. motor is more efficient if it runs more slowly?

Whether or not the motor is under load, the key electrical equation is $$\mathscr E_{supply}-\mathscr E_{back} =IR.$$ This provides clear-cut answers to your questions...
No. When the motor is first connected to the power supply the back-emf is zero because the conductors aren't moving and cutting flux. So the current and therefore the magnetic force, $$BI\ell,$$ are high. Eventually the motor reaches a speed at which the back-emf almost cancels the power supply emf, so the back-emf is high and the current and magnetic force are low.
To study efficiency under load, it helps to multiply the above equation through by I: $$\mathscr E_{supply}I-\mathscr E_{back}I =I^2R$$ $$\mathscr E_{supply}I$$ is the power input and $$I^2R$$ is the power dissipated in the coils of the motor. Because of energy conservation we must identify $$\mathscr E_{back}I$$ with mechanical output power, $$G \omega.$$ This is useful output power, so long as we neglect resistive torque in the bearings and brush/commutator rub. So I don't go along with your "Given the smaller back E.M.F. and therefore smaller energy loss [...]" Indeed the motor efficiency is $$\eta=\frac{G \omega}{\mathscr E_{supply} I}=\frac{\mathscr E_{back} I}{\mathscr E_{supply} I}=\frac{\mathscr E_{back}}{\mathscr E_{supply}}$$