Does Schroedinger equation depend on the sign of Poisson bracket? Let's consider Poisson bracket
$$\left\{ A,B\right\} =\alpha_{p} \left( 
\frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q^{k}}-\frac{%
\partial A}{\partial q^{k}}\frac{\partial B}{\partial p_{k}}\right) \tag{1}$$
where $\alpha _{p}=\pm 1;$ then
$$\frac{dA}{dt}=\frac{1}{%
\alpha_{p}}\left\{ H,A\right\} \tag{2}$$
we know relation between Poisson bracket and commutator 
$$\left\{ \hat{O}_{A},\hat{O}_{B}\right\} =i \alpha_{c}\left[
\hat{O}_{A},\hat{O}_{B}\right] \tag{3}$$
then Heisenberg equation
$$\frac{d\hat{A}}{dt}=\frac{i \alpha_{c}%
}{\alpha_{p} }\left[ \hat{H},\hat{A}\right] \tag{4}$$
for momentum and coordinate
$$\frac{d\hat{p}}{dt}=\frac{i \alpha_{c}}{\alpha_{p} }\left[ \hat{H},\hat{p}\right] \qquad\text{and} \qquad\frac{d\hat{q}}{dt}=\frac{i \alpha_{c}}{\alpha_{p} }\left[ \hat{H},\hat{q}%
\right] \tag{5}$$
finally we get
$$\frac{i \alpha_{c}}{\alpha_{p} }\frac{\partial \psi 
}{\partial t}=H\psi\tag{6} $$
so there are two independent parameters $\alpha_{c}$ and $\alpha_{p} $ I can choose $\alpha_{c}=1$ and $\alpha_{p}=-1$.
Is that computation correct?
 A: *

*With the standard CCR
$$[\hat{q},\hat{p}]~=~i\hbar{\bf 1},  \tag{A}$$
and with OP's convention (1) for the Poisson bracket $$\{p,q\}_{PB}~=~\alpha_p,\tag{B}$$
the operator $\leftrightarrow$ function correspondence reads
$$ [\hat{f},\hat{g}]\quad\longleftrightarrow\quad\frac{\hbar}{i\alpha_p}\{f,g\}_{PB} +{\cal  O}(\hbar^2) .\tag{C}$$ 
Note that the traditional physics convention is $\alpha_p=-1$. See e.g. this related Phys.SE post.

*Now let's address OP's title question. The time-dependent Schrödinger equation (TDSE)
$$ i\hbar \frac{d |\psi \rangle}{d t}~=~\hat{H}|\psi \rangle,   \tag{D}$$ 
and the Heisenberg equation
$$ \frac{d\hat{f}}{dt}~=~\frac{i}{\hbar}[\hat{H},\hat{f}] + \frac{\partial \hat{f}}{\partial t},   \tag{E}$$ 
are independent of the Poisson bracket convention (1);
while Hamilton's/Liouville's equation
$$ \frac{df}{dt}~=~\frac{1}{\alpha_p}\{H,f\}_{PB} + \frac{\partial f}{\partial t},  \tag{F}$$ 
does depend on the Poisson bracket convention (1).

*For discussions of non-standard sign conventions in the TDSE, see e.g. this & this Phys.SE posts.
A: Because you add a sign to the Poisson brackets, you should do the same to the commutator for the relation
$$\left\{\hat{O}_{A},\hat{O}_{B}\right\}\leftrightarrow i\left[\hat{O}_{A},\hat{O}_{B}\right]$$
to still hold. Thus you just define the Poisson brackets/commutator differently, keeping the explicit equations the same. There is no point of doing this.
EDIT 1: The time evolution of a quantity in analytical mechanics is unique
$$\frac{dA}{dt}=\left\{A,H\right\}$$
with the regular definition of the Poisson brackets. Also Ehrenfest's theorem is unique
$$\frac{d\left<\hat{A}\right>}{dt}=\frac{1}{i\hbar}\left<\left[\hat{A},\hat{H}\right]\right>$$
with the regular definition of the commutator. This leads us to the usual correspondence
$$\left\{A,B\right\}\leftrightarrow\frac{1}{i\hbar}\left[\hat{A},\hat{B}\right]$$
If for some reason you want to change the usual definitions and write
$$\left\{A,B\right\}_{\rm OP}=\alpha_{\rm PB}\left\{A,B\right\}$$
$$\left[\hat{A},\hat{B}\right]_{\rm OP}=\alpha_{\rm C}\left[\hat{A},\hat{B}\right]$$
then all the above equations change accordingly, but that's only because you defined something differently. It is only a cosmetic change.
