If the vector $(u,v)$ is independent of the vector $x$, then I would like to show that $$E(u|x,v)= E(u|v)$$
The only thing I can derive from the definitions is that if $(u,v)$ is independent of $x$, then $E( (u,v) | x)= E((u,v))$.
I can no longer attack this problem!
Help
$E(u|v)$ is measurable w.r.t $\sigma (x,v)$ so we only have to show that $$EI_{x^{-1}(A)\cap v^{-1}(B)} E(u|x,v)=EI_{x^{-1}(A)\cap v^{-1}(B)}E(u|v)$$ for all Borel sets $A,B$ in $\mathbb R$.
Left side is $$EI_{x^{-1}(A)} [I_{v^{-1}(B)}] E(u|x,v)].$$ By the assumed independence this becomes $$P(X^{-1}(A))E[I_{v^{-1}(B)}] E(u|x,v)].$$ Finally we can write this as $P(x^{-1}(A))E[I_{v^{-1}(B)}u]$.
Right side is $$EI_{x^{-1}(A)}I_{v^{-1}(B)}E(u|v)=P(x^{-1}(A)) EI_{v^{-1}(B)}E(u|v)$$ because $I_{v^{-1}(B)}E(u|v)$ is independent of $x$. So the left side is also equal to $$P(x^{-1}(A))E[I_{v^{-1}(B)}u].$$