My intuition on this says, no, since the $\chi^2$ random variable $U=\sum_{i=1}^k Z_i^2$ where each $Z_i$ is an independent Standard Normal random variable.
That said, I was told by a professor some time ago that the sample mean is independent from the sample variance of a Normal distribution, so does it follow that U is independent from Z?
Your professor's comment is about Basu's theorem, see the Example section also. As for your question, which is also counter-intuitive, think about the case where $k=1$, i.e. $U=Z^2$. A simple proof of dependence would be to show a non-zero covariance value, however these variables have zero covariance. Another one is to think of conditional distribution $f_{U|Z}(u|z)$. Independence forces us to have $$f_{U|Z}(u|z)=f_U(u)$$ which is not correct since $f_{U|Z}(u|z)$ have a degenerate PDF which is equal to $\delta(u-z^2)\neq f_U(u)$, not a Chi-Squared distribution. Intuitively, if we know $Z=z$, then $U$ will certainly be $u=z^2$ without any uncertainty left.
