We assume that the short interest rate $r_t$ follows the Hull-White model, that is, the short rate $r$ and the stock price $S$ satisfies a system of SDEs of the form
\begin{align*}
dr_t &= (\theta_t -a\, r_t)dt + \sigma_0 dW_t,\\
dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t + \sqrt{1-\rho^2} dB_t\Big)\Big],
\end{align*}
where $a$, $\sigma_0$, $\sigma$, and $\rho$ are constants, and $\{W_t, t\ge 0\}$ and $\{B_t, t\ge 0\}$ are two independent standard Brownian motions.

Note that,
\begin{align*}
&\ E\bigg(\exp\Big(-\int_0^T r_t dt \Big) (S_T-K)^+\bigg) \\
=& \  E\bigg(e^{-\bar{r}T} \Big(S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T - \sigma (\rho W_T + \sqrt{1-\rho^2}B_T)} -K\Big)^+ \bigg)\\
=& \ E\Bigg(E\bigg(e^{-\bar{r}T} \Big(S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T + \sigma (\rho W_T + \sqrt{1-\rho^2}B_T)} -K\Big)^+ \mid r_s, 0<s \leq T\bigg)\Bigg)\\
=& \ E\Big(F(S_0,K,\bar{r},T,\sigma, W_T) \mid r_s, 0<s \leq T\Big),
\end{align*}
for a certain function $F$. Note the random variable $W_T$ in the formula.

If $\rho=0$, that is, $S$ and $r$ are independent, then
\begin{align*}
&\ E\bigg(\exp\Big(-\int_0^T r_t dt \Big) (S_T-K)^+\bigg) \\
=& \ E\Bigg(E\bigg(e^{-\bar{r}T} \Big(S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T + \sigma B_T} -K\Big)^+ \mid r_s, 0<s \leq T\bigg)\Bigg)\\
=&\ E\Big(BS(S_0,K,\bar{r},T,\sigma) \mid r_s, 0<s \leq T \Big).
\end{align*}
That is, the formula provided in the question holds if the stock price and the interest rate are independent. In this case, $\bar{r}$ can be approximated by a Riemann sum.

>EDIT

Here, we provide an analytical valuation formula for the above vanilla European option. From [this question][1],  the zero-coupon bond price is given by
\begin{align*}
P(t, T) &=  E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right)\\
&=\exp\left(-B(t, T) r_t - \int_t^T  \theta(s) B(s, T) ds + \frac{1}{2}\int_t^T \sigma_0^2 B(s, T)^2 ds\right),
\end{align*}
where
\begin{align*}
B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big).
\end{align*}
Then 
\begin{align*}
d\ln P(t, T) &=-e^{-a(T-t)}r_tdt -B(t, T)dr_t + \theta(t)B(t, T)dt - \frac{1}{2} \sigma_0^2 B(t, T)^2 dt\\
&=\left(r_t-\frac{1}{2} \sigma_0^2 B(t, T)^2\right) dt - \sigma_0 B(t, T)dW_t,\tag{1}
\end{align*}
or 
\begin{align*}
dP(t, T) = P(t, T)\big[r_t dt -  \sigma_0 B(t, T)dW_t\big].
\end{align*}


Let $Q$ denote the risk-neutral measure and $Q^T$ denote the $T$-forward measure. Moreover, let $B(t) = e^{\int_0^t r_s ds}$ be the money market account value. From $(1)$,
\begin{align*}
\frac{dQ^{T}}{dQ}\big|_t &= \frac{P(t, T)}{P(0, T)B(t)}\\
&=\exp\left(-\frac{1}{2}\int_0^t \sigma_0^2 B(s, T)^2 ds - \int_0^t \sigma_0 B(s, T) dW_s\right).
\end{align*}
Then, under $Q^T$, the process $\{(\widehat{W}_t, \widehat{B}_t), t \ge 0 \}$, where
\begin{align*}
\widehat{W}_t &= W_t + \int_0^t \sigma_0 B(s, T) ds,\\
\widehat{B}_t &= B_t,
\end{align*}
is a standard two-dimensional Brownian motion. Moreover, under $Q^T$,
\begin{align*}
dP(t, T) &= P(t, T)\big[r_t dt -  \sigma_0 B(t, T)dW_t\big]\\
&=P(t, T)\Big[\big(r_t +\sigma_0^2 B^2(t, T)\big)dt -  \sigma_0 B(t, T)d\widehat{W}_t\Big]\\
dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t + \sqrt{1-\rho^2} dB_t\Big)\Big]\\
&=S_t\Big[\big(r_t- \rho\sigma_0\sigma B(t, T)\big)  dt + \sigma \Big(\rho d\widehat{W}_t + \sqrt{1-\rho^2} d\widehat{B}_t\Big)\Big].\tag{2}
\end{align*}


Note that, the forward price $F(t, T)$ has the form
\begin{align*}
F(t, T) &= E_{Q^T}(S_T \mid \mathcal{F}_t)\\
&=\frac{S_t}{P(t, T)}.
\end{align*}
which is a martingale under the $T$-forward measure $Q^T$ and satisfies an SDE of the form
\begin{align*}
dF(t, T) &= \frac{dS_t}{P(t, T)} -\frac{S_t}{P^2(t, T)}dP(t, T) \\
&\qquad - \frac{\langle dS_t, dP(t, T)\rangle}{P^2(t, T)} + \frac{S_t}{P^2(t, T)}\langle dP(t, T), dP(t, T)\rangle\\
&= F(t, T)\left[\sigma \Big(\rho d\widehat{W}_t + \sqrt{1-\rho^2} d\widehat{B}_t\Big) + \sigma_0 B(t, T)d\widehat{W}_t \right]\\
&= F(t, T) \left[ \big(\sigma\rho + \sigma_0 B(t, T)\big)  d\widehat{W}_t + \sigma \sqrt{1-\rho^2} d\widehat{B}_t \right].
\end{align*}
Let $\hat{\sigma}$ be a quantity defined by
\begin{align*}
T\hat{\sigma}^2 &= \int_0^T\Big[\big(\sigma\rho + \sigma_0 B(t, T)\big)^2 + \sigma^2\big(1-\rho^2\big) \Big] ds\\
&=\int_0^T\Big[\sigma^2 + 2\rho\sigma\sigma_0 B(t, T) +  \sigma_0^2 B^2(t, T)\Big] ds\\
&=\sigma^2T + \frac{2\rho\sigma\sigma_0}{a}\Big[T-\frac{1}{a}\big(1-e^{-aT}\big)\Big] + \frac{\sigma_0^2}{a^2}\Big[T+\frac{1}{2a}\big(1-e^{-2aT} \big) - \frac{2}{a}\big(1-e^{-aT} \big) \Big]\\
&=\sigma^2T + \frac{2\rho\sigma\sigma_0}{a}\Big[T-\frac{1}{a}\big(1-e^{-aT}\big)\Big] + \frac{\sigma_0^2}{a^2}\Big[T-\frac{1}{2a}e^{-2aT}+\frac{2}{a}e^{-aT} -\frac{3}{2a} \Big].
\end{align*}
Then
\begin{align*}
F(T, T) = F(0, T)\exp\left(-\frac{1}{2}\hat{\sigma}^2T + \hat{\sigma}\sqrt{T} Z \right),
\end{align*}
where $Z$ is a standard normal random variable. Consequently,
\begin{align*}
E_Q\left(\frac{(S_T-K)^+}{B(T)}\right) &= E_Q\left(\frac{(F(T, T)-K)^+}{B(T)}\right)\\
&=E_{Q^T}\left(\frac{(F(T, T)-K)^+}{B(T)} \left(\frac{dQ^{T}}{dQ}\big|_T\right)^{-1} \right)\\
&=P(0, T)E_{Q^T}\left((F(T, T)-K)^+\right)\\
&=P(0, T)\big[F(0, T)N(d_1) - KN(d_2) \big],
\end{align*}
where  $d_1 = \frac{\ln F(0, T)/K + \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}$ and $d_2 = d_1 - \hat{\sigma} \sqrt{T}$. 


[1]: http://quant.stackexchange.com/questions/21513/extended-hull-white-interest-rate-model-for-zero-coupon-bond/21516#21516.