Let's see how we can play with Noether's theorem in the "conservation of energy" example.
First of all, we apply a time-independent variation to the Lagrangian,
\begin{equation}
t \rightarrow t+ \epsilon
\end{equation}
For a general Lagrangian, it will change
\begin{equation}
L(x(t), \dot{x}(t), t)\rightarrow L'(x(t), \dot{x}(t), t ) = L(x(t), \dot{x}(t), t-\epsilon )
\end{equation}
what we require in this case, is that the Lagrangian is invariant(not just covariant) as a scalar,
\begin{equation}
L=L' \implies \frac{\partial L}{\partial t} = 0
\end{equation}
That's a property true alone any path.
Then we use an induced time-dependent variation,
\begin{equation}
t \rightarrow t+ \epsilon(t) \quad x(t) \rightarrow x(t) + \epsilon(t) \dot{x(t)}
\end{equation}
The change of the action in this case is
\begin{equation}
\delta S = \int_0 ^T \frac{\partial L }{\partial x} \dot{x(t)} \epsilon(t) + \frac{\partial L } {\partial \dot{x} } \frac{d}{dt}(\epsilon(t) \dot{x}(t)) = \int_0^T dt \epsilon(t) ( \frac{\partial L }{\partial x} \dot{x} + \frac{\partial L } {\partial \dot{x} } \ddot{x} ) + \frac{\partial L } {\partial \dot{x} } \dot{x} \dot{\epsilon}
\end{equation}
Now adding $\frac{\partial L}{\partial t}=0$ to the first two terms in the parentheses to make it a total derivative, since this is true for all paths in the configuration space . For the second term, we integrate by part to get(variation has zero boundary condtions),
\begin{equation}
\delta S = \int_0^T dt \epsilon(t) \frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x})
\end{equation}
Alone the classical path, any variation will extremize the action; in particular our special variation $\epsilon(t)$ induced by the symmetry action will. We get the energy conservation equation,
\begin{equation}
\frac{d}{dt}(L - \frac{\partial L } {\partial \dot{x} } \dot{x}) = 0 \quad \text{along classical path}
\end{equation}
In other words, you have to choose $T$ and $Q$ to be constant(independent of time) to dig out the symmetries of the Lagrangian. Then you can apply the time-dependent variation to get a special "equation of motion" along the classical path: conservation law.
I need to elaborate the symmetry equation I have derived.
Under a time-independent variation, we say Lagrangian actually changes
\begin{equation}
L' = L(x(t-\epsilon), \dot{x}(t-\epsilon), t-\epsilon )
\end{equation}
This is due to an identity,
\begin{equation}
L(x(t),\dot{x}(t),t) = L'(t') \quad t'=t+\epsilon
\end{equation}
That means the value of Lagrangian of a point in configuration space is independent of the time coordinate describing it. It's like you can use different time zones to describe an event. They only differ by a constant.
However, this is not true if we apply a time dependent variation,
\begin{equation}
\frac{d}{dt}(x(t-\epsilon)) = \frac{d}{dt}( x(t) - \epsilon \dot{x}(t) ) = \dot{x}(t) - \dot{\epsilon}\dot{x(t)} - \epsilon \ddot{x}(t) \ne \dot{x}(t) - \epsilon \ddot{x}(t) = \dot{x}(\tau)|_{\tau = t-\epsilon}
\end{equation}
the time dependent variation forces the velocity to change in a another way different from what we like. It's like time is not flowing uniformly, such that even if you displace your velocity vector to the original point before the translation, it still changes, because the rate will depend on how time flows.
In the technical aspect, you variation can't lead to
\begin{equation}
\frac{\partial L}{\partial t} = 0
\end{equation}
which is equivalent to the conservation law.
