Energy of a damped oscillator $$
E=\frac{1}{2}m\left(\frac{dx}{dt}\right)^2+\frac{1}{2}m\omega_0^2x^2.
$$
This is the equation for the energy of a oscillator. The second term is the potential energy. 
Now, my question is, will this second term remain the same if we are to consider the oscillator as a damped oscillator.
Because doesn't the second term arise from the consideration of a conservative force field? And when talking of damped oscillator, isn't the force field non-conservative?
 A: Yes, that equation will still give the correct value for the energy of the oscillator system at any point in time, assuming of course that you know dx/dt and x at that time.   If there is an external dissipative force on the system (damping) you will find that the value of E decreases with time.  But the energy of the oscillator itself is still the sum of the kinetic energy, and the potential energy associated with the restoring force.  The effect of the damping is to remove energy from the oscillator.
One could ask about a different system:  that consisting of the oscillator, plus the "sink" where the damped energy ends up (usually increased thermal energy of some part of the environment).  In that case, the expression for the total energy of the system will have a term that accounts for the increase of thermal energy in the "sink".  The value of that energy will remain constant.
A: Your second term is still referring to the conservative force acting on the object. For an ideal spring, this would mean: $\omega_0^2=\frac{k}{m}$. However, in the case of a damped oscillator, there will be both conservative and not conservative forces acting on the object with mass $m$.
But this does not mean that the potential energy of the conservative force will be lost. This can still be transferred to kinetic energy or dissipated by the non-conservative forces.
