Let $\nu_x$ be the regular conditional probability associated with $X$, and $\mu_X$ the law of $X$ on $\mathbb R^n$.

Denote by $E$ the event 
$$\left \{ \nu_X ( \bigcap_i \, \{X^i \in A_i\} ) =
\prod_i \nu_X (X^i \in A_i) \, , \, \forall A_i \in \mathcal B(C[0, T])\right \}.$$

By definition of conditional independence, we need to show that $\mathbb P(E) = 1.$

But for $\mu_X$-almost every $x$, the $X^i_0$ are deterministic under $\nu_x$, and hence also the process $\eta_s$. As such, each $X^i$ is a standard diffusion SDE driven by $B^i$ with non-random coefficients, for which it is known there is a strong solution. 

Thus there exists some deterministic map $F$ such that each $X^i = F(B_i)$ almost surely under $\nu_x$ for $\mu_X$-almost every $x$, and hence the independence of the $X^i$ for $\mu_X$-almost every $x$ follows from that of the $B_i$.

In other words, denoting by $S$ the set


$$\{ x \in \mathbb R^n \, | \, \nu_x ( \bigcap_i \, \{X^i \in A_i\} ) =
\prod_i \nu_x (X^i \in A_i) \, , \, \forall A_i \in \mathcal B(C[0, T]) \}$$

we have $\mu_X (S) = 1$, and so

$$\mathbb P (E)$$
$$= \int_{\mathbb R^n} 1_S (X(\omega)) \, d\mathbb P (\omega)$$
$$= \int_{\mathbb R^n} 1_S (x) \, d\mu_X (x)$$
$$ = 1.$$

And so we conclude conditional independence of the processes $X^i$ as desired.