The following claim characterizes circumstances under which $\ker(\text{eval}_q)$ for $q \in K = \text{Frac}(R)$ is a principal ideal of $R[X]$. In particular, this answers question $2$ in the **positive** and provides a recipe to construct non-principal ideals of the form $\ker(\text{eval}_q)$.

**Claim 1.** Let $R$ be an integral domain and let $q = \frac{a}{b} \in \text{Frac}(R)$ . Then the following are equivalent:

$(i)$ The ideal of denominators of $q$, that is $\mathfrak{d} \Doteq \{d \in R \,\vert\, dq \in R\}$, is the principal ideal of $R$ generated by $b$.

$(ii)$ For every $c \in R$, if $b$ divides $ac$ then $b$ divides $c$, in other words $Ra \cap Rb = Rab$.

$(iii)$ The ideal $\ker(\text{eval}_q)$ is the principal ideal of $R[X]$ generated by $bX - a$.

**Edit:** This characterization is well-known under a slightly different form, see Addendum at the bottom.

Note that if $a, b \in R$ are *co-prime* in the sense that $Ra + Rb = R$, then $(ii)$ is satisfied. If $R$ is a Prüfer domain, e.g., a Dedekind domain, the condition $(ii)$ is equivalent to $Ra + Rb = R$. The condition $\text{gcd}(a, b) = 1$ doesn't imply $(ii)$ in general, but If $R$ is a pre-Schreier domain, then $(ii)$ is equivalent to $\text{gcd}(a, b) = 1$. You can find in [1] an example of a Schreier domain which is not a GCD domain.

*Proof of Claim 1.* $(i) \Rightarrow (ii)$ is immediate. Let us show that $(ii) \Rightarrow (iii)$. To do so, let us consider $f(X) = \sum_{i = 0}^n a_i X^i \in \ker(\text{eval}_q)$ with $n > 0$. There is $g(X) = \sum_{i = 0}^{n - 1} b_i X^i \in K[X]$ such that $f(X) = (bX - a)g(X)$. We deduce from the latter identity that $b_{n - 1 - i} = \frac{ab_{n - i} + a_{n - i}}{b}$ for every $0 \le i \le n - 1$, agreeing that $b_n = 0$. The identity $f(q) = 0$ can be re-written as $$a_n a^n + a_{n- 1}a^{n - 1}b + \cdots + a_1 ab^{n - 1} + a_0b^n = 0.$$ Hence it follows from $(ii)$ that $b$ divides $a_n$, so that the above equality is equivalent to
$$
a^{n - 1}(ab_{n - 1} + a_{n - 1}) + a_{n - 2}a^{n - 2}b + \cdots + a_1 ab^{n - 2} + a_0b^{n - 1 } = 0
$$
where $b_{n - 1} = \frac{a_n}{b} \in R$. Substituting $ab_{n - 1} + a_{n - 1}$ with $bb_{n -2}$, dividing the left-hand side by $b$ and using $(ii)$ again yields $b_{n - 2} \in R$. By repeating this process we eventually obtain that $g(X) \in R[X]$. Therefore $f(X) \in R[X](bX - a)$, which establishes $(iii)$.
We will complete the proof of Claim $1$ by showing that $\neg (i) \Rightarrow \neg (iii)$. By hypothesis we can find $d \in \mathfrak{d}$ such that $b$ doesn't divide $d$. As result $dX - dq \in \ker(\text{eval}_q) \setminus R[X](bX - a)$.

The next claim underlines the special rôle of GCD domains with respect to OP's questions.

**Claim 2.**
Let $R$ be an integral domain. Then the following are equivalent:

$(i)$ For every $q \in \text{Frac}(R)$, the ideal of denominators of $q$ is a principal ideal of $R$.

$(ii)$ $R$ is a GCD domain.

If $q = \frac{a}{b}$, then $\mathfrak{d} = \frac{1}{a}(Ra \cap Rb)$. Thus the ideal of denominators of $q$ is principal if and only $Ra \cap Rb$ is principal, that is $\text{lcm}(a,b)$ exists. In this case, we know that $\text{gcd}(a, b)$ exists and is such that $\text{lcm}(a,b)\text{gcd}(a, b) = ab$ up to multiplication by a unit.

*Proof of Claim 2.* Assertion $(i)$ is equivalent to the fact that $\text{lcm}(a,b)$ exists for every $a, b \in R$. The latter is equivalent to the fact that $\text{gcd}(a,b)$ exists for every $a, b \in R$.

The following consequence is immediate.

**Corollary.** Let $R$ be an integral domain. Then the following are equivalent:

$(i)$ For every $q \in \text{Frac}(R)$, the ideal $\ker(\text{eval}_q)$ is a principal ideal of $R[X]$.

$(ii)$ $R$ is a GCD domain.

In general, we observe that $\{ bX - a \,\vert \, q = \frac{a}{b} \} = \{ dX - dq \,\vert \, d \in \mathfrak{d} \setminus \{0\}\}$ is the set of polynomials in $\ker(\text{eval}_q)$ with degree $1$. The question as to whether the latter set generates $\ker(\text{eval}_q)$ was answered in the **negative** by a user44191's comment: this doesn't hold when $R = \mathbb{Z}[\sqrt{5}]$ and $q = \frac{1 + \sqrt{5}}{2} \in \text{Frac}(R)$. Clearly, the ideal $\mathfrak{d}$ of denominators of $q$ contains $2$ and $1 - \sqrt{5}$. It is easy to check that the ideal $\mathfrak{m} = R \cdot 2 + R \cdot (1 - \sqrt{5})$ is a non-principal maximal ideal of $R$ of index $2$. Thus $\mathfrak{d} = \mathfrak{m}$ is not principal. As $X^2 - X -1 \in \ker(\text{eval}_q)$, any polynomial in $\ker(\text{eval}_q)$ is of the form $f(X)(X^2 - X - 1) + g(X)$ with $f(X), g(X) \in R[X]$, $g(X) = 0$ or $\deg(g(X)) \le 1$. Therefore $$\ker(\text{eval}_q) = R[X]((1 - \sqrt{5})X + 2) + R[X]((2X - (1 + \sqrt{5})X) + R[X](X^2 - X - 1).$$

But we can get a **positive answer** for a class of rings which are not necessarily Noetherian. A domain $R$ is called a *locally GCD domain* if every localization $R_{\mathfrak{m}}$ of $R$ at a maximal ideal $\mathfrak{m}$ of $R$ is a GCD domain.

**Claim 3.** Let $R$ be a locally GCD domain, e.g., a Prüfer domain, and let $q \in \text{Frac}(R)$. Then $\ker(\text{eval}_q)$ is generated as an ideal of $R[X]$ by the set $\{ bX - a \,\vert \, q = \frac{a}{b} \}$.

I ignore if the converse of Claim 3 holds true.

*Proof of Claim 3.* Let $i \ge 1$ and let $\mathfrak{c}_i$ be the ideal of $R$ generated the leading coefficients of the polynomials in $\ker(\text{eval}_q)$ with degree $i$. We have in particular $\mathfrak{c}_1 = \mathfrak{d}$, the ideal of denominators of $q$. We observe first that $\ker(\text{eval}_q)$ is generated by polynomials of degree at most $n \ge 1$ if and only if $\mathfrak{c}_i = \mathfrak{c}_n$ for every $i > n$. We shall establish that $\mathfrak{c}_i = \mathfrak{c}_1$ for every $i > 1$. Let $\mathfrak{m}$ be a maximal ideal of $R$ and let $\ker(\text{eval}_q)_{\mathfrak{m}}$ be the kernel of the evaluation map $f \mapsto f(q)$ from $R_{\mathfrak{m}}[X]$ to $\text{Frac}(R)$. Since $\mathfrak{c}_i = \left( \frac{b}{a^i} (Ra + Rb)^{i - 1} \right)\cap R$, the ideal generated by the leading coefficients of the polynomials in $\ker(\text{eval}_q)_{\mathfrak{m}}$ with degree $i$ is $\mathfrak{c}_i R_{\mathfrak{m}}$. As $R_{\mathfrak{m}}$ is a GCD domain by hypothesis, we deduce from Claim 2, that $\mathfrak{c}_i R_{\mathfrak{m}} = \mathfrak{c}_1 R_{\mathfrak{m}}$ for every $i > 1$. As this holds for every maximal ideal $\mathfrak{m}$, we deduce that $\mathfrak{c}_i = \mathfrak{c}_1 $ holds for every $i > 1$.

**Addendum.**
I discovered that Claim 1 above is known under the following form:

**Claim 4 [2, Exercise 17.2]**. Let $R$ be a commutative integral domain and $a \in R, b \in R \setminus \{0\}$ and let $q = \frac{a}{b} \in F(R)$. Then the following are equivalent:

- $(i)$ The sequence $(b, a)$ is a regular.
- $(ii)$ The cohomology group $H^1(K(b,a))$ of the Koszul complex of $(b, a)$ is trivial, that is $(Rb: Ra)/Rb = \{0\}$.
- $(iii)$ The polynomial $(bX - a)$ is a prime element of $R[X]$.
- $(iv)$ The ideal $\ker(\text{eval}_q)$ is the principal ideal of $R[X]$ generated by $bX - a$.

A sequence $(a_1, \dots, a_n)$ of elements in a ring $R$ is said to be *regular* if for each $i$ the element $a_i$ is a regular element of $R/(Ra_1 + \cdots Ra_{i - 1})$. Given two ideals $I, J$ of $R$, we used above the following notation: $(I : J) \Doteq \{ r \in R \,\vert\, rJ \subseteq I \}$.

[1] P. M. Cohn, "Bezout rings and their subrings", 1968.

[2] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.