Let $G \in \mathcal{F}_r$. Using that
$$1_G \int_r^t u_s \, dB_s = \int_r^t 1_G u_s \, dB_s \quad \text{a.s.} $$
(see the lemma below) it follows from Itô's isometry that
$$\begin{align*} \int_G \left( \int_r^t u_s \, dB_s \right)^2 \, d\mathbb{P} &= \int \left( \int_r^t u_s 1_G \, dB_s \right)^2 \, d\mathbb{P} \\ &= \mathbb{E} \left( \left| \int_r^t u_s 1_G \, dB_s \right|^2 \right) \\ &= \int_r^t \mathbb{E}(u_s^2 1_G) \, ds \\ &= \int_G \left( \int_r^t u_s^2 \, ds \right) \, d\mathbb{P}. \end{align*}$$
As $G \in \mathcal{F}_r$ is arbitrary, this shows that $$\mathbb{E} \left( \left[ \int_r^t u_s \, dB_s \right]^2 \mid \mathcal{F}_r \right) = \mathbb{E} \left( \int_r^t u_s^2 \, ds \mid \mathcal{F}_r \right).$$
Lemma For any $G \in \mathcal{F}_r$ it holds that $$1_G \int_r^t u_s \, dB_s = \int_r^t 1_G u_s \, dB_s \quad \text{a.s.} \tag{1}$$
Proof: It is straight-forward to check that $(1)$ holds for simple functions $u$, i.e. functions of the form $$u(s) = \sum_{j=1}^n \xi_{j-1} 1_{[t_{j-1},t_j)}(s)$$
where $r \leq t_0 < \ldots < t_n \leq t$ and $\xi_{j-1} \in L^2(\mathcal{F}_{t_{j-1}})$. Now if $u$ is an $\mathcal{F}$-adapted process such that $\int_0^t \mathbb{E}(u_s^2) \, ds < \infty$, then there exists a sequence of simple functions $(u^{(n)})_{n \geq 1}$ such that $\int_0^t \mathbb{E}(|u^{(n)}(s)-u(s)|^2) \, ds \to 0$ as $n \to \infty$ and $\int_r^t u^{(n)}_s \, dB_s \to \int_r^t u_s \, dB_s$ in $L^2$. Then it follows (e.g. from Itô's isometry) that $\int_r^t u^{(n)}_s 1_G \, dB_s \to \int_r^t u_s 1_G \, dB_s$. Using that the assertion holds for each $u^{(n)}$ we obtain $$\begin{align*} 1_G \int_r^t u_s \, dB_s &= L^2-\lim_{n \to \infty} 1_G \int_r^t u^{(n)}_s \, dB_s \\ &= L^2-\lim_{n \to \infty} \int_r^t 1_G u^{(n)}_s \, dB_s \\ &= \int_r^t 1_G u_s\, dB_s. \end{align*}$$
Best Answer
This equality cannot hold true as the following counterexample shows:
For fixed $t>0$ set $X_s(\omega) := 1_{[0,t]}(s)$. Then the left-hand side is equal to
$$\mathbb{E} \bigg( \underbrace{\left(\int_0^t X_s \, dB_s \right)^2}_{B_t^2} \mid \mathcal{F}_t^B \bigg) = \mathbb{E}(B_t^2 \mid \mathcal{F}_t^B)=B_t^2$$
whereas the right-hand side equals
$$\mathbb{E} \left(\int_0^t X_s^2 \, ds \mid \mathcal{F}_t^B \right) = \mathbb{E}(t \mid \mathcal{F}_t^B)=t$$