Itô's lemma to solve the SDE Given $dG_{t}=\alpha S_{t}dt+\upsilon S_{t}dW_{t}$ and $dS(t)={dG_{t}}-\epsilon_{t}dt$.
How can I have $S_{t}=\mathbb{E}^{\mathbb{Q}}\left[\int_{t}^{+\infty}e^{-r(s-t)}\epsilon_{t}ds|\mathcal{F}_{t}\right]$ where $W_{t}^{\mathbb{Q}}=W_{t}+\int_{0}^{t}\frac{\alpha-r}{\upsilon}ds$
 A: I didn't understood from where comes $r$ which intervenes at your change of mesure.
We can easelly show by Ito's Lemma that $\forall \lambda \> 0$, the following processs 
$$ Z_t = \exp \left\{ \int_0 ^t  \lambda_s dW_s -\frac {1}{2} \int _0 ^t (\lambda_s )^2 ds \right\}, \ \ t\geq 0$$
is a $\mathcal F_t$- martingal. 
So, we can define the probability mesure, equivalent to $\mathbb P$, $ \mathbb Q$ by
$$ d \mathbb Q_{\mathcal F_t} = Z _td \mathbb P_{\mathcal F_t}$$ 
and have a $\mathbb Q $- brownien motion given by 
 $$ W^{\mathbb Q} _ t  =  W_t  -\int_0 ^t \lambda_s  ds $$
In your problem $\lambda_s  = \frac{r-\alpha}{v} : = \lambda$ , where $r$, $\alpha$ and $v$ must be constants, I suppose. So,  $\int_0^t \lambda_s ds = \frac{r-\alpha}{v} t$ and 
\begin{align} Z_t &= \exp \left\{  \frac{r-\alpha}{v}W_t -\frac {1}{2} (\frac{r-\alpha}{v} )^2 t \right\}, \ \ t\geq 0 \\ &= \exp \left\{ \lambda W_t -\frac {1}{2} (\lambda)^2 t \right\}, \ \ t\geq 0\end{align}
then, 
\begin{align} \mathbb E^{\mathbb{Q}} \left \{  \int_0^t e^{-r(s-t)} \epsilon_t ds  | \mathcal F _t\right \} &= \mathbb E^{\mathbb{P}} \left \{ Z_t \int_0^t e^{-r(s-t)} \epsilon_t ds  | \mathcal F _t\right \}  \\ &= \mathbb E^{\mathbb{P}} \left \{  \int_0^t e^{\lambda W_t - \frac{1}{2}\lambda^2 t-r(s-t)} \epsilon_t ds  | \mathcal F _t\right \}  \\ &= (...)\end{align}
wich should lead to the result using the hypothesys about the dynamics of both $S_t$ and $G_t$. Please verifie also if it's $\mathbb E^{\mathbb{Q}} \left \{  \int_0^t e^{-r(s-t)} \epsilon_s ds  | \mathcal F _t\right \}$ or  $\mathbb E^{\mathbb{Q}} \left \{  \int_0^t e^{-r(s-t)} \epsilon_t ds  | \mathcal F _t\right \}$ as you wrote.
