The answer to all three questions is "no".

This answer uses the same strategy as my earlier answer, but is a lot simpler in details because it puts most of the complexity into the ring $S$ rather than the ideal $I$. Let $k$ be an infinite field.
Let $p_1$, $p_2$, ..., $p_9$ be $9$ generic points in $\mathbb{P}_k^2$; let $x$, $y$ and $z$ be the homogenous linear coordinates on $\mathbb{P}^2$. Let $A = k[w_1, w_2, \ldots, w_9,x,y,z]$.
We will consider $A$ as a ring graded by $\mathbb{Z}^{10}$, where we separately keep track of the degrees in the $w_j$'s but lump the $x$, $y$ and $z$ degrees together.
So an element of $A$ is homogenous if it is of the form $\prod_i w_i^{r_i} \cdot f(x,y,z)$ with $f$ a homogenous polynomial of some degree, which we will denote $d$.
All our rings will be graded subrings of $A$.

Let $N_9$ be the graded subring of $A$ where $\prod w_i^{r_i} \cdot f(x,y,z)$ is in $N_9$ if, for each $i$, the polynomial $f(x,y,z)$ vanishes to order $\geq d-r_i$ at $p_i$. Let $N_8$ be the analogous subring of $k[w_1, \ldots, w_8, x,y,z]$, using the points $p_1$ through $p_8$. Nagata showed that $N_8$ is finitely generated over $k$, and $N_9$ is not.

Let $S = N_8[w_1]$. Choose coordinates on $\mathbb{P}^2$ so that $x$ and $y$ vanish at $p_9$ and $z$ does not. Set $\tilde{x} = w_1 w_2 \cdots w_8 x$, $\tilde{y} = w_1 w_2 \cdots w_8 y$ and $\tilde{z} = w_1 w_2 \cdots w_8 w_9 z$. Set $R = k[w_1, w_2, \ldots, w_9, \tilde{x}, \tilde{y}, \tilde{z}]$. So $R \subset S$; as $S$ is a finitely generated $k$ algebra, it is a finitely generated $R$ algebra.

Set $I = w_9 R$. We claim that
$$S \cap R_I = N_9. \quad (\ast)$$
Furthermore, we claim that the kernel of $N_9 \to R_I/I R_I$ is $w_9 N_9$. So the image is $N_9/(w_9 N_9)$, which we will show below is also not finitely generated.

Since $S$ is a graded subring of $A$, as is $R$, and $I$ is a graded ideal, the intersection $S \cap R_I$ is a graded subring.
Let $g=\prod w_i^{r_i} f(x,y,z)$ be a homogenous element, with $\deg f =d$.

First, suppose that $g \in N_9$. Then clearly $g \in S$. Also, since $f$ vanishes to order $\geq d-r_9$ at $p_9$, we can write $w_9^{r_9} f(x,y,z)$ as a polynomial in $x$, $y$ and $z w_9$. Note that $x$, $y$ and $z$ are $(w_1 \cdots w_8)^{-1} \tilde{x}$, $(w_1 \cdots w_8)^{-1} \tilde{y}$ and $(w_1 \cdots w_8)^{-1} \tilde{z}$ so they are in $R_I$. Therefore, $w_9^{r_9} f(x,y,z)$ is in $R_I$. Since $w_1$ through $w_8$ are also in $R_I$, this shows that $g$ is in $R_I$. So $N_9 \subseteq S \cap R_I$.

Now, suppose that $g$ is in $S \cap R_I$. Since $g \in S$, we see that $f$ vanishes to order $\geq d-r_i$ at $p_i$ for $1 \leq i \leq 8$.
Since $g$ in $R_I$, we can write $g=a/b$ with $a \in R_I$ and $b \in R_I - I$. Since $R$ is a UFD, we can assume that $a$ and $b$ are relatively prime in
$R = k[w_1, w_2, \ldots, w_9, \tilde{x}, \tilde{y}, \tilde{z}]$. If $b$ has factors other than the $w_j$'s, then $a/b$ won't be in $R$, sos we may assume that $b$ is of the form $\prod_{j \leq 8} w_j^{q_j}$.
We can then write $g = \prod_{j \leq 8} w_j^{s_j} \cdot w_9^{s_9} h(\tilde{x}, \tilde{y}, \tilde{z}) = \prod_{j \leq 8} w_j^{s_j+d} \cdot w_9^{s_9} h(x,y,w_9 z)$ for some constants $s_j$.
If we write $h(x,y,w_9 z) = w_9^r f(x,y,z)$ then $f$ vanishes to order $d-r$ at $p_9$ and $d-r \geq d-(r+s_9)=d-r_9$.
So we have shown that $g$ is in $N_9$. We have now established $(\ast)$.

Similar arguments show that $g \in S \cap I R_I$ if and only if $f$ vanishes to order $> d-r_9$ at $p_9$, if and only if $w_9$ divides $g$ in $N_9$.
I'll omit the details (but please challenge me if you think I'm wrong!).

So the image of $N_9$ in $R_I / I R_I$ is $N_9 / w_9 N_9$. We now recall the following lemma:

**Lemma:** Let $N$ be a positively graded $k$-algebra whose degree $0$ part is $k$. Let $J$ be a finitely generated graded ideal of $N$. Then $N/J$ is finitely generated if and only if $N$ is.

**Proof:** Clearly, if $N$ is finitely generated, then any quotient of $N$ is. Conversely, suppose that $a_1$, $a_2$, ..., $a_m$ is a list of generators for $J$ as an ideal, and $\bar{b}_1$,
..., $\bar{b}_n$ is a list of generators for the $k$-algebra $N/J$. Let $b_i$ be a lift of $\bar{b}_i$ to $N$.
We may assume that the $a_i$ and $b_i$ are homogenous and of positive degree. We claim that $N$ is generated by the $a$'s and $b$'s.
We show, by induction on $N_d$, that $N_d$ is contained in the $k$-subalgebra of $N$ generated by the $a$'s and $b$'s.
Let $g \in N_d$. Then there is a polynomial $p$ so that $g \equiv p(a_1, \ldots, a_m) \bmod J$. Thus, $g = p(a_1, \ldots, a_m) + \sum b_i h_i$ for some $h_i \in N$. The $h_i$ are of lower degree, so inductively, the $h_i$ are in the subalgebra generated by the $a$'s and $b$'s, and thus $g$ is in this algebra. $\square$

Since $N_9$ is not finitely generated, this lemma shows that $N_9 / w_9 N_9$ isn't either.