Stumbled by coincidence over this question. Therefore the late answer.

In the graded case it is true that $k[x_1,...,x_n]$ is integral over $k[f_1,...,f_n]$ ($k$ a field). But even more is true:

Let $k$ be a field and let $R = \oplus_{i\ge 0}R_i $ be a finitely generated graded commutative $k$-algebra with $R_0= k$. Assume moreover that $R$ is Cohen-Macaulay and that
$y_1,...,y_n$ is a regular sequence in $R$. Set $I := \oplus_{i> 0}R_i$. Then

$\sqrt{I} = (y_1,...,y_n)$

There is $k > 0$ with $I^k \subseteq (y_1,...,y_n)$.

$R$ is a free $k[y_1,...,y_n]$-module.

Proof: 2) It's well-known that if $y \in R$ is a homogeneous regular element of positive degree, then Krull-dim $R/(y) =$ Krull-dim$(R) - 1$. Applying this repeatedly to $R/(y_1,...,y_i)$ one obtains that $\bar{R} := R/(y_1,...,y_n)$ has dimension $0$. Hence $\bar{R}$ is Artinian and thus there is $k > 0$ such that $\bar{R}_i = 0$ for all $i \ge k$. So, if $x \in I$ then $\bar{x}^k = 0$ in $\bar{R}$, i.e. $x^k \in (y_1,...,y_n)$.

1) Follows from 2)

3) Induction on $n$. Case $n=0$: $R$ is Artinian and hence a finite dim. $k$ vector space.
Assume the case $n-1$ is true. Since $\tilde{R} := R/(y_1)$ is again CM and $\tilde{y}_2,...,\tilde{y}_n$ a regular sequence, there are $\tilde{a}_i \in \tilde{R}$ with $\tilde{R} = \oplus_i k[\tilde{y}_2,...,\tilde{y}_n]\tilde{a}_i$.

Let's show $R = \oplus_i k[y_1,...,y_n]a_i$: Let $f_i \in k[y_1,...,y_n]$ with $\sum_i f_i a_i = 0$.

We proceed by induction on $m := \max_i\; \deg_{y_1}(f_i)$: Reduction modulo $y_1$ shows $f_i \in (y_1)$ for all $i$. If $m=0$, then $f_i = 0$ for all $i$. If $m> 0$ let $f'_i = f_i/x_1$. Thus $y_1\sum_i f'_i a_i = 0$ and since $y_1$ is regular, we conclude $\sum_i f'_i a_i = 0$ with $m' < m$. Then by induction hypothesis $f'_i = 0$ and hence $f_i = 0$.
q.e.d.