One application of Girsanov's theorem is in stochastic control theory. Suppose I have uncontrolled process $dX_t = \sigma^0(X_t)dW^0_t + dW^1_t$ under $\mathbb P$ on $[0, T]$, where $W^0$ and $W^1$ are independent $\mathbb P$-standard BMs. Let's say $\sigma^0$ is Lipschitz with linear growth to guarantee a unique strong solution.
Given some control $\alpha_{t \in [0, T]}$ progressively measurable w.r.t. the two BMs, and a bounded controlled drift function $b(t, x, a)$, I can define new probability measure $\mathbb{P}^{\alpha}$ by Girsanov change of measure using stochastic exponential $$\frac{d\mathbb{P}^{\alpha}}{d\mathbb{P}} := \mathcal{E}\left(\int_0^\cdot b(t, X_t, \alpha_t)dW^1_t \right) \mid_T, \quad W^{\alpha}_t := W^{1}_t - b(t, X_t, \alpha_t)dt.$$
Girsanov's theorem tells me that $(W^0, W^\alpha)$ is a 2-dim standard BM under $\mathbb{P}^{\alpha}$, and $X$ solves the following controlled SDE under $\mathbb{P}^\alpha$: $$dX_t = b(t, X_t, \alpha_t)dW^\alpha_t + \sigma(X_t)dW^0_t.$$
Now suppose that $b$ is also nice (Lipschitz, linear growth, bounded, etc) so that this controlled SDE is uniquely solvable under $\mathbb{P}$ in the first place, i.e. I have $$d\widetilde{X}_t = b(t, \widetilde{X}_t, \alpha_t)dW^1_t + \sigma(\widetilde{X}_t)dW^0_t.$$ Then in particular, we have $\mathbb{E}^{\mathbb{P}^{\alpha}}[X_t] = \mathbb{E}^{\mathbb{P}}[\widetilde{X}_t]$ because they share the same law under respective probability measures.
My question is whether I can also say $\mathbb{E}^{\mathbb{P}^{\alpha}}[X_t|\mathcal{F}^0_s] = \mathbb{E}^{\mathbb{P}}[\widetilde{X}_t|\mathcal{F}^0_s]$ (almost surely in what sense?) for $s \leq t$ where $\mathcal{F}^0$ is the filtration generated by $W^0$.
If so, then I can take Malliavin derivative to both, w.r.t. $W^0$ and move it into the conditional expectation. However, the driftless $X$ is Malliavin differentiable but $\widetilde{X}$ is not, since it is controlled. I believe the answers to these two questions are "Yes" and "No", which puzzles me.
Much appreciated!