In order to accomodate generalized coordinates, suppose that the Lagrangian has the form $$ L(q, \dot q, t)=\frac{1}{2}m_{ij}(q)\dot{q}^i\dot{q}^j-U(q,\dot{q},t). $$
The first term is also denoted $T(q,\dot{q})$. Let $\delta_i:=\delta/\delta q^i$ denote the Euler--Lagrange derivative. Then $$ \delta_i T=-m_{ik}(q)\left(\ddot{q}^k+\Gamma^{k}_{jl}(q)\dot{q}^j\dot{q}^l\right)=-m_{ik}A^k, $$ where $A^k:=\ddot{q}^k+\Gamma^k_{ij}(q)\dot{q}^i\dot{q}^j$, while $$ \delta_i U=\frac{\partial U}{\partial q^i}-\frac{d}{dt}\frac{\partial U}{\partial \dot{q}^i}=-Q_i, $$ where $Q_i$ is the so-called generalized force. So the EL equation $\delta_iL\approx 0$ is equal to the generalized Newton equation $$ Q_i = m_{ij}A^j,\qquad Q_i=-\delta_iU. $$
1. Classical potentials do not necessarily imply conservativity.
Let $U=U(q,t)$, i.e. it does not depend on the velocity. Then $\delta_i U=\partial U/\partial q^i$. Let $$ \frac{d L}{dt}=\frac{\partial L}{\partial t}+\delta_i L\dot{q}^i + \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}^i}\dot{q}^i\right), $$ so if we set $$ E(q,\dot{q},t):=\frac{\partial L}{\partial \dot{q}^i}\dot{q}^i-L(q,\dot{q},t)=\frac{1}{2}m_{ij}(q)\dot{q}^i\dot{q}^j+U(q,t), $$ whenever the equations of motion hold, we have $$ \frac{dE}{dt}=-\frac{\partial L}{\partial t}=\frac{\partial U}{\partial t}, $$ i.e. the total energy is not conserved, but the forces are still derived from a potential, as $Q_i=-\partial U/\partial q^i$.
Conservation of energy also necessitates that $U=U(q)$, i.e. the potential does not depend on $t$ explicitly.
2. The case with velocity dependent potentials.
First note that for Newton-like equations where the velocity-dependent potential $U(q,\dot{q},t)$ is to be interpreted as a potential, the EL derivative $\delta_i U$ should not depend on $\ddot{q}^i$. This implies the constraint $$ U(q,\dot{q},t)=V(q,t)+W_i(q,t)\dot{q}^i, $$ i.e. it is an affine function of the velocity, so we have a scalar potential $V$ and a vector potential $W_i$. This will lead to a Lorentz-like force. The energy function now takes the form $$ E(q,\dot{q},t) = \frac{1}{2}m_{ij}(q)\dot{q}^i\dot{q}^j+ V(q,t) $$ again, i.e. the vector potential does not contribute to the energy function. We still have the relation $$ \frac{dE}{dt}=-\frac{\partial L}{\partial t}=\frac{\partial U}{\partial t}=\frac{\partial V}{\partial t} + \frac{\partial W_i}{\partial t}\dot{q}^i. $$
So the energy function is conserved on solutions if and only if both $V$ and $W_i$ are time-independent, i.e. $U=U(q,\dot{q})=V(q)+W_i(q)\dot{q}^i$. In this case we can still talk about a conservative system, since energy is conserved.
