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^1,\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big)\Big], \end{align*} where $a$, $\sigma_0$, $\sigma$, and $\rho$ are constants, and $\{W_t^1, t\ge 0\}$ and $\{W_t^2, t\ge 0\}$ 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 \big(\rho W_T^1 + \sqrt{1-\rho^2}W_T^2\big)} -K\Big)^+ \bigg)\\ =& \ E\Bigg(E\bigg(e^{-\bar{r}T} \Big[S_0e^{\bar{r}T -\frac{1}{2}\sigma^2 T + \sigma \big(\rho W_T^1 + \sqrt{1-\rho^2}W_T^2\big)} -K\Big]^+ \Bigg\vert r_s, 0<s \leq T\bigg)\Bigg)\\ =& \ E\Big(F(S_0,K,\bar{r},T,\sigma, W_T^1) \Big\vert r_s, 0<s \leq T\Big), \end{align*} for a certain function $F$. Note the random variable $W_T^1$ 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 W_T^2} -K\Big)^+ \bigg\vert r_s, 0<s \leq T\bigg)\Bigg)\\ =&\ E\Big(BS(S_0,K,\bar{r},T,\sigma) \Big\vert 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, the zero-coupon bond price is given by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \Big\vert \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*}\begin{align*} \frac{dP(t, T)}{P(t, T)} = r_t dt - \sigma_0 B(t, T)dW_t. \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}\Bigg|_t &= \frac{P(t, T)B_0}{P(0, T)B_t}\ \ (\text{with } B_0=1) \\ &=\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 by the Girsanov theorem, under $Q^T$, the process $\{(\widehat{W}_t^1, \widehat{W}_t^2), t \ge 0 \}$, where \begin{align*} \widehat{W}_t^1 &= W_t^1 + \int_0^t \sigma_0 B(s, T) ds,\\ \widehat{W}_t^2 &= W_t^2, \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^1\big]\\ &=P(t, T)\Big[\big(r_t +\sigma_0^2 B(t, T)^2\big)dt - \sigma_0 B(t, T)d\widehat{W}_t^1\Big]\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big)\Big]\\ &=S_t\Big[\big(r_t- \rho\sigma_0\sigma B(t, T)\big) dt + \sigma \Big(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2\Big)\Big].\tag{2} \end{align*}\begin{align*} \frac{dP(t, T)}{P(t, T)} &= r_t dt - \sigma_0 B(t, T)dW_t^1 \\ &=\big(r_t +\sigma_0^2 B(t, T)^2\big)dt - \sigma_0 B(t, T)d\widehat{W}_t^1 \\ \frac{dS_t}{S_t} &= r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big) \\ &=\big(r_t- \rho\sigma_0\sigma B(t, T)\big) dt + \sigma \Big(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2\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(t, T)^2}dP(t, T) \\ &\qquad - \frac{d\langle S_t, P(t, T)\rangle}{P(t, T)^2} + \frac{S_t}{P(t, T)^3}d\langle P(t, T), P(t, T)\rangle\\ &= F(t, T)\left[\sigma \Big(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2\Big) + \sigma_0 B(t, T)d\widehat{W}_t^1 \right]\\ &= F(t, T) \left[ \big(\sigma\rho + \sigma_0 B(t, T)\big) d\widehat{W}_t^1 + \sigma \sqrt{1-\rho^2} d\widehat{W}_t^2 \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(s, T)\big)^2 + \sigma^2\big(1-\rho^2\big) \Big] ds\\ &=\int_0^T\Big[\sigma^2 + 2\rho\sigma\sigma_0 B(s, T) + \sigma_0^2 B^2(s, 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} \frac{dQ}{dQ^T}\bigg|_T \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 $d_2 = d_1 - \hat{\sigma} \sqrt{T}$.
