The answer of VA is quite simple to understand. In fact the result is of local nature.

<b>Proposition</b>: Let $R$ be a noetherian (commutative unitary) ring, let $B$ be a flat noetherian $R$-algebra, and let $f\in B$ be an element such that for any maximal ideal $m$ of $R$, the image of $f$ in $B/mB=B\otimes_R R/m$ is a regular element. Then $B/fB$ is flat over $R$. 

This can be found for example in Matsumura, page 151, (20.F) (taking $M=B$). It is also in Milne's "Etale cohomology", first chapter. 

To apply to your concrete situation, $B$ is a polynomial ring over $R$, and $f$ is a polynomial which is non-zero modulo $m$ (cf. explanation of VA), so it is regular modulo $m$ because $B/mB$ is an integral domain. Therefore $B/fB$ is flat over $R$.  

<b>Edit</b> As pointed out by @DamienRobert in the comment, for any maximal ideal $\mathfrak n$ of $B$, we need $f$ to be regular in $B/{\mathfrak p}B$ where $\mathfrak p$ is the pre-image of $\mathfrak n$ in $R$.  It would be safe to require in the proposition that the pre-image of a maximal ideal of $B$ is maximal (for example if $A$ is Jacobson and $B$ is finite type over $A$), or ask $f$ to be regular in the quotients $B/{\mathfrak p}B$ for all prime ideals $\mathfrak p$ of $A$. 

Still I don't have a counterexample without these assumptions.