If $\mathcal A$ is generated by global sections and $k$ is infinite then a regular section must exist. For each component of $X$, global sections vanishing on (if necessary, the induced reduced subscheme of) that component form a positive-codimension subspace of global sections. Over an infinite field, we may avoid arbitrarily many positive codimension subspaces with a single vector.

If $k$ is finite then this is not true. We could take $X$ to be the union of all $\mathbb F_q$-rational lines in $\mathbb P^2$ and $\mathcal A = \mathcal O(1)$. Then using the ideal sheaf of $X$ we can check that all global sections of $\mathcal A$ are the obvious global sections of $\mathcal O(1)$ on $\mathbb P^2$, each of which vanishes on a line.

But probably, in the situation you're thinking of, it's fine to assume $k$ is infinite.

If $\mathcal A$ is generated by global sections and $k$ is infinite then a regular section must exist. For each irreducible component of $X$, global sections vanishing on that component form a positive-codimension subspace of global sections. For non-reduced $X$, global sections vanishing on the induced reduced subscheme of any associated prime also find a positive-codimension subspace. 

Over an infinite field, we may avoid arbitrarily many positive codimension subspaces with a single vector.

If $k$ is finite then this is not true. We could take $X$ to be the union of all $\mathbb F_q$-rational lines in $\mathbb P^2$ and $\mathcal A = \mathcal O(1)$. Then using the ideal sheaf of $X$ we can check that all global sections of $\mathcal A$ are the obvious global sections of $\mathcal O(1)$ on $\mathbb P^2$, each of which vanishes on a line.

But probably, in the situation you're thinking of, it's fine to assume $k$ is infinite.