Summary
I want to clarify how can I prove the fact that "the Noether charge generates the corresponding transformation" when the infinitesimal transformation of the fields contain the canonical momentum $\pi$.
Formulation
Let us consider the transformation of a set of scalar fields $\phi^i(x)$ on $D$-dimensional Minkowski space whose metric is $(- + + \dots +)$, \begin{align} \delta_\epsilon \phi^i (x) = \epsilon^A (\Delta_A \phi)^i (x). \end{align} Here, we simply assume the transformation corresponds to a Lie algebra and $A$ denotes its indices. $\Delta_A$ is usually assumed to be a linear oprator which includes some partial derivatives with spacetime $x$ and matirces acting on $i$ like the flavor symmetry.
For the moment, we assume there exists a Lagrangian of the system and it does not depend on time explicitly.
The condition which assures that the transformation is indeed the symmetry: $\delta \mathcal{L} = \epsilon^A \partial_\mu {K^{\mu}}_A$.
Then we can define the Noether charge as \begin{align} Q_A &:= - \int d^{D-1} \vec{x} ~ \left( \frac{\delta \mathcal{L}}{\delta \partial_0 \phi} (\Delta_A \phi)^i - K^0_A\right)\\ &= - \int d^{D-1} \vec{x} ~ \left( \pi_i (\Delta_A \phi)^i - K^0_A\right), \end{align} as displayed in Supergravity by Daniel Z. Freedman and Antoine Van Proeyen.
It is well-known fact that the Noether charge generates the corresponding symmetry as \begin{align} (\Delta_A \phi)^i = \{Q_A, \phi^i \}_{{\rm P.B.},~t=t_0} \tag{1}, \end{align} with the canonical commutation relation assumed.
In the following, I assume the usual Lagrangian $\mathcal{L} = -\frac{1}{2} \partial_\mu \phi^i \partial^\mu \phi ^i +V(\phi)$, which gives the canonical momentum as \begin{align} \pi_i = \partial_0 \phi. \end{align}
Then, if $K^0_A = 0$ and $\Delta_A \phi$ does not contain $\partial_0 \phi$, it can be proven easily.
When $\Delta_A \phi$ contains $\pi$
However, I do not understand how should we treat $\Delta_A \phi$ when it contains the time derivative $\partial_0 \phi$.
For the time translation, we can simply replace $\partial_0 \phi \in \delta_A \phi$ by $\pi$;
\begin{align} \{Q_A, \phi \} = -\int d^{D-1} \vec{x} \left\{ {\frac{1}{2}\pi_i \pi^i}, \phi^i \right\} \end{align}
which gives a correct result.
Hence, I think that it is actually switched to the Hamilton formalism with $(\phi, \pi)$, when I write down the definition of the Noether charge.
(a) However, is that replacement valid for arbitary cases where $\delta_A \phi$ contains $\partial_0 \phi$?
(b) How can I prove eq. (1) in such cases ($K^0_A$ is in general nonzero, and $\{\pi \Delta_A \phi, \phi \} \neq \{\pi , \phi \}\Delta_A \phi$)?
