Does $X\perp\!\!\!\perp Y\implies\mathbb{E}[Z|X]\perp\!\!\!\perp\mathbb{E}[Z|Y]$ for any r.v. Z?
I would think so, because $X\perp\!\!\!\perp Y$ implies $\sigma(X)\perp\!\!\!\perp\sigma(Y)$ and $\sigma(\mathbb{E}[Z|X])\subset\sigma(X)$ as well as $\sigma(\mathbb{E}[Z|Y])\subset\sigma(Y)$ for any r.v. $Z$ per definition, q.e.d.
However my intuitions fail half the time in probability theory, so I would be happy if someone could confirm this or point out an error.
Yes! Perhaps the easiest way to see this is as a result of the Doob-Dynkin Lemma, which states that if $X$ is a $\sigma(Y)$-measurable random variable, where $\sigma(Y)$ is the $\sigma$-algebra generated by $Y$, then $X=g(Y)$ for some borel-measurable function $g: \mathbb{R} \mapsto \mathbb{R}$. In your case, since $E[Z|X]$ is $\sigma(X)$-measurable, $E[Z|X]=g_1(X)$, and similarly $E[Z|Y]=g_2(Y)$. Now the problem reduces to showing that if $X$ and $Y$ are independent, then $g_1(X)$ and $g_2(Y)$ are independent for Borel-mesaurable $g_1$ and $g_2$. This is a straightforward exercise to establish.