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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 $bX - a$.

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 $dX - dq \in \ker(\text{eval}_q) \setminus R[X](bX - a)$.

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

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 $a, b \in R$.

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

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 $\{ bX - a \,\vert \, q = \frac{a}{b} \}$.

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 $i > 1$.

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

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 $dX - dq \in \ker(\text{eval}_q) \setminus R[X](bX - a)$.

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

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 $a, b \in R$.

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

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 $\{ bX - a \,\vert \, q = \frac{a}{b} \}$.

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.

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 $\mathfrak{c}_i = \left( \frac{b}{a^i} \sum_{j = 1}^{i} Ra^{i - 1}b^{j - 1} \right)\cap R$, 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 $i > 1$.

If $R$ is a number ring, I can only prove the following at the moment:

Claim 4. Let $R$ be an order in a quadratic number field. Then the following are equivalent:

$(i)$ The ideal $\ker(\text{eval}_q)$ is generated by polynomials of degree $1$.
$(ii)$ The ideal $\ker(\text{eval}_q)$ is principal.
$(iii)$ The element $q$ is of the form $\frac{a}{b}$ for some integer $a$ of $\text{Frac}(R)$ and some rational integer $b > 1$ which is co-prime with $N_{\mathbb{Q}}(a)$, the norm of $a$ over $\mathbb{Q}$.

I'll add the details later.

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 $i > 1$.

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 prime 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 $\mathfrak{c}_i = \left( \frac{b}{a^i} \sum_{j = 1}^{i} Ra^{i - 1}b^{j - 1} \right)\cap R$, 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 $i > 1$.

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 $\mathfrak{c}_i = \left( \frac{b}{a^i} \sum_{j = 1}^{i} Ra^{i - 1}b^{j - 1} \right)\cap R$, 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 $i > 1$.