We define a sequence of (discrete) stopping times
$$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$
It is not difficult to see that $\tau_j$ is indeed a stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Since the Brownian motion has continuous paths, this implies $B(\tau) = \lim_{j \to \infty} B(\tau_j)$.
Let $\xi,\eta \in \mathbb{R}$. Then, by the dominated convergence theorem,
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \lim_{j \to \infty} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau_j+t)-B(\tau_j))} \cdot e^{\imath \, \eta B(\tau_j)} \bigg) \\ &= \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \xi (B(k \cdot 2^{-j} +t)-B(k \cdot 2^{-j}))} \cdot e^{\imath \, \eta B(k \cdot 2^{-j})} \cdot 1_{\{\tau_j = k \cdot 2^{-j}\}} \bigg) \end{align*}$$
where we used in the last step that $\tau_j$ is a discrete stopping time. By assumption, $B(k \cdot 2^{-j}+t)-B(k \cdot 2^{-j})$ and $B(k \cdot 2^{-j}) \cdot 1_{\{\tau_j=k2^{-j}\}}$ are independent. Therefore, we obtain
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \eta B(k 2^{-j})} \cdot 1_{\{\tau_j=k \cdot 2^{-j}\}} \bigg) \\ &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg). \end{align*}$$
(In the second step we used again dominated convergence, similar to the above calculation.) If we choose $\eta = 0$, then we get
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \bigg) = \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg);$$
hence,
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg)= \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg)$$
i.e. $B(\tau+t)-B(\tau)$ and $B(\tau)$ are independent. Therefore, the strong Markov property gives
$$\begin{align*} \mathbb{E}\bigg(1_F e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid \mathcal{F}_{\tau} \bigg] \bigg) \\ &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid B_{\tau} \bigg] \bigg)\\ &= \mathbb{P}(F) \cdot \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \end{align*}$$
for any $F \in \mathcal{F}_{\tau}$. Consequently, $B(\tau+t)-B(\tau)$ is independent of $\mathcal{F}_{\tau}$.
A very similar calculation shows that
$$\mathbb{E} \left( \exp \left( \imath \sum_{j=1}^n \xi_j \cdot (B(\tau+t_j)-B(\tau+t_{j-1})) \right) \right) = \prod_{j=1}^n \mathbb{E}e^{\imath \, \xi_j B(t_j-t_{j-1})}$$
for any $\xi_j \in \mathbb{R}$, $0 \leq t_0 < \ldots \leq t_n$. This means that $(B(\tau+t_j)-B(\tau+t_{j-1}))_{j=1,\ldots,n}$ are independent normal distributed random variables.
From $2$ to $1$:
Let $V=(V_t) = (B_{T+t} - B_t)$ be that process, let $A \in \mathcal B(\mathbb R^{[0,\infty)})$ (in cylinder $\sigma-$field). Let $A_0 = \{ x \in \mathbb R^{[0,\infty)} : x-x(0) \in A \}$ ( we translate every function in $A$ by minus it's value at $0$). Note that:
$$ \mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_x( (B_{T+t} - B_T) \in A | \mathcal F(T)) = \mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T))$$
(since $B_T$ is value at $0$ of process $(B_{T+t})_{t \ge 0}$. Now apply $2$, getting:
$$ \mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T)) = \mathbb P_{B_T}( (B_t) \in A_0) $$
Note that for any $y \in \mathbb R$ we have:
$$ \mathbb P_y( (B_t) \in A_0 ) = \mathbb P_y( (B_t - y) \in A) = \mathbb P_0 ( (B_t) \in A) $$
Taking $y = B_T$ it finally gives us:
$$ \mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_0 ( (B_t) \in A) $$
in particular $$ \mathbb P_x (V \in A) = \mathbb E_x[\mathbb P_x(V \in A |\mathcal F(T))]= \mathbb P_0( ( B_t) \in A) $$
so we showed that $V$ has the same distribution (under $\mathbb P_x$ measure) as standard brownian motion $(B_t)$ (cause it's under $\mathbb P_0$ measure))
Now to show independence, Take any $B \in \mathcal F(T)$ we get:
$$ \mathbb P_x(B \cap \{ V \in A\}) = \mathbb E_x [ 1_B \mathbb P_0 ( (B_t) \in A)) = \mathbb P_x(B)\mathbb P_0( (B_t) \in A) = \mathbb P_x(B)\mathbb P_x( V \in A) $$
From 1 to 2:
$$\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_x [ f(B_{T+t} - B_{T} + B_{T}) | \mathcal F(T)] $$
I don't know what information you possess, but it can be shown that as adapted, right continuous process, Brownian Motion is progresivelly measurable, hence $B_T$ is $\mathcal F(T)$ measurable. By 1. we have that $B_{T+t} - B_T$ is independent of $\mathcal F(T)$ so by conditional expected value property, the last one is equal to:
$$ \mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T} $$
Again using $1$, we know that $B_{T+t} - B_T$ under $\mathbb P_x$ is distributed as standard (so under $\mathbb P_0$) brownian motion, so:
$$ \mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T} = \mathbb E_0 [ f(B_t + p)]|_{p = B_T} = \mathbb E_p[f(B_t)]|_{p =B_T} = \mathbb E_{B_T}[f(B_t)]$$
So we proved $\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_{B_T}[f(B_t)]$.
Best Answer
No. Note that the stopping time $T$ can depend on all the values of $B_t$ with $t\le T$. So you could take something like $$T=\inf\{t\ge 1: B_t=0, B_{t-1/2}>0\}.$$ In this case $(B_T-B_{T-t})_{t\in [0,1]}$ cannot be a Brownian motion, since its value at $t=1/2$ is always negative.