Assume you have a probability space $(\Omega,\mathcal{A},P)$. Assume you have a sub sigma-algebra and two random variables $X,Y$ such that $E(|X|), E(|Y|), E(|XY|)<\infty$. Do we then have
$$E(X|\mathcal{G})\cdot E[Y|\mathcal{G}]=E[XY|\mathcal{G}]?$$
Both sides are $\mathcal{G}$-measurable, so we only need to check the integrability requirement. So let $A\in \mathcal{G}$, what we need to show is that
$$E[E(X|\mathcal{G})\cdot E[Y|\mathcal{G}]\cdot1_A]=E[XY\cdot 1_A].$$
Do you see if this holds?
No. Let $B_t$ be Brownian motion and $\mathcal{F}_t$ be its natural filtration. Set $X = Y = B_t$. Observe that $B_t$ and $B_t^2 - t$ are martingales. Then, for $s \leq t$, $$E(X | \mathcal{F}_s)E(Y | \mathcal{F}_s) = B_s^2$$
But, $$E(XY |\mathcal{F}_s) = E(B_t^2 \, | \, \mathcal{F}_s) = E(B_t^2 - t\, | \, \mathcal{F}_s) + t = B_s^2 -s + t$$