The geometric interpretation of Noether's normalization lemma is that any affine algebraic variety has a finite surjective morphism to the affine space $\mathbb A^d_k$ of dimension $d=\dim X$. When $k$ is an integral domain, ** and $X$ dominates ** $\mathrm{Spec}(k)$, the finite surjective morphism of the generic fiber of $X$ to $\mathbb A^d_{K}$, where $K=\mathrm{Frac}(k)$ extends to a finite surjective morphism $X_V\to V$ for some dense open subset $V$ of $\mathrm{Spec}(k)$. This is the geometric interpretation of the statement in M. Hochster's note in Harry Gindi's post.

In general, such a morphism can not exist because it would imply that the fibers of $X\to \mathrm{Spec}(k)$ all have the same dimension.

Suppose that this condition is satisfied: the fibers of $X\to\mathrm{Spec}(k)$ all have the same dimension $d$. Then I think that a reasonable statement (Noether's normalization lemma over a ring $k$) would be: there exists a quasi-finite and surjective morphism $X\to \mathbb{A}^d_k$. If $k$ is noetherian, then by Zariski's Main Theorem, this implies that $X$ is an open subscheme of scheme which is finite surjective over $\mathbb{A}^d_k$. In general one can not expect better result than quasi-finite (consider the case $d=0$).

The above "reasonable statement" should be easy to prove when $k$ is a local ring.