Combining insight from Mahdi's and Yusuf's answers, it looks as thought the finite map in question is virtually never flat.

Specifically, assume $R$ is an integral domain, and $K = K(R)$ is its fraction field. Let $a \in K$ be integral over $R$. Consider the finite morphism $$\operatorname{Spec} R[a] \to \operatorname{Spec} R.$$ Since $R[a]$ and $R$ have the same fraction field $K$, this morphism has degree one (is birational). If it is flat, then every fiber must have the same degree (one); in other words, this morphism is an isomorphism in every fiber. Finiteness, together with Nakayama's Lemma, then implies it is an isomorphism above every stalk. Hence, it is an isomorphism, i.e., $a \in R$.

Contrapositively, if $a \not\in R$, then the morphism is not flat.

For a specific example suggested by Yusuf's answer, consider $R = \Bbbk[x^2, x^3]$, and $a = x = x^3 / x^2 \in K(R)$.

Combining insight from Mahdi's and Yusuf's answers, it looks as thought the finite map in question is virtually never flat.

Specifically, assume $R$ is an integral domain, and $K = K(R)$ is its fraction field. Let $a \in K$ be integral over $R$. Consider the finite morphism $$\operatorname{Spec} R[a] \to \operatorname{Spec} R.$$ Since $R[a]$ and $R$ have the same fraction field $K$, this morphism has degree one (is birational). If it is flat, then every fiber must have the same degree (one); in other words, this morphism is an isomorphism in every fiber. Finiteness, together with Nakayama's Lemma, then implies it is an isomorphism above every stalk. Hence, it is an isomorphism, i.e., $a \in R$.

Contrapositively, if $a \not\in R$, then the morphism is not flat.

For a specific example suggested by Yusuf's answer, consider $R = \Bbbk[x^2, x^3]$, and $a = x = x^3 / x^2 \in K(R)$.