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
$$\mathbb{E} \left(\int_0^t X_s^2 \, ds \mid \mathcal{F}_t^B \right) = \mathbb{E}(t \mid \mathcal{F}_t^B)=t$$
Actually, the proof is indeed similar to the proof of $(1)$. It's based on the fact that convergence in probability implies almost sure convergence of a subsequence. Set
$$\tau_n := \inf\left\{t \geq 0; \int_0^t |\sigma(s)|^2 \, ds + \int_0^t |b(s)|^2 \, ds \geq n \right\}.\tag{4}$$
By Doob's inequality, Itô's isometry and Tschbysheff inequality we have
$$\begin{align} & \quad \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^t (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon \right) \\ &\leq \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^{t \wedge \tau_n} (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon, \tau_n > T \right) + \mathbb{P}(\tau_n \leq T) \\ &\leq \frac{4}{\varepsilon^2} \cdot \underbrace{\mathbb{E} \left( \int_0^{T \wedge \tau_n} |f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)|^2 \, ds \right)}_{=:I} + \mathbb{P}(\tau_n \leq T) \end{align} $$
(Note that the boundedness of $f'$ implies the existence of the integrals.) Since \begin{align*} f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s) &= f'(X_s^{\Pi}) \cdot \big(\sigma^{\Pi}(s)-\sigma(s)\big) \\ &\quad + \sigma(s) \cdot \big(f'(X_s^{\Pi})-f'(X_s) \big)\end{align*} and $(a+b)^2 \leq 2a^2+2b^2$ we obtain $$\begin{align*} I &\leq 2 \mathbb{E} \bigg( \int_0^{T \wedge \tau_n} \underbrace{|f'(X_s^{\Pi})|^2}_{\leq \|f'\|^2_{\infty}} \cdot |\sigma^{\Pi}(s)-\sigma(s)|^2 \, ds \bigg)\\ &\quad + 2 \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \end{align*}$$
The first addend converges to $0$ as $|\Pi| \downarrow 0$ since $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$ by assumption. For the second one, we note that
\begin{align*} \int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \, ds &\leq \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2 \int_0^{\tau_n} |\sigma(s)|^2 \, ds \\ &\leq n \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2 \end{align*}
Since $X^{\Pi} \to X$ uniformly in probability, it follows from the continuity of $f'$ that the right-hand side converges to $0$ in probability as $|\Pi| \to 0$. Moreover, by the above estimate,
$$\int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \,d s \leq 2n \|f'\|_{\infty} \in L^1(\mathbb{P})$$
and so Vitali's convergence theorem gives
$$ \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \to 0$$
Similarily, one can prove the convergence of the other addends in the right-hand side of $(2)$.
Best Answer
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 $$\mathbb{E} \left( \left[ \int_r^t u_s \, dB_s \right]^2 \mid \mathcal{F}_r \right) = \mathbb{E} \left( \int_r^t u_s^2 \, ds \mid \mathcal{F}_r \right).$$
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 $$\begin{align*} 1_G \int_r^t u_s \, dB_s &= L^2-\lim_{n \to \infty} 1_G \int_r^t u^{(n)}_s \, dB_s \\ &= L^2-\lim_{n \to \infty} \int_r^t 1_G u^{(n)}_s \, dB_s \\ &= \int_r^t 1_G u_s\, dB_s. \end{align*}$$