If you take a mapping class $\psi:\Sigma\to \Sigma$ whose characteristic polynomial of its action on homology $\psi_\ast: H_1(\Sigma;\mathbb{Z}) \to H_1(\Sigma;\mathbb{Z})$ is irreducible, then it cannot extend over a handlebody. The point is that if $\psi= \Psi_{| \Sigma}$, where $\Psi: H\to H, \partial{H}=\Sigma$, $H$ a handlebody, then $\psi$ preserves $K=ker\{H_1(\Sigma)\to H_1(H)\}$, which is an integral Lagrangian subspace of $H_1(\Sigma)$. So the characteristic polynomial of $\psi_{\ast | K}$ divides the characteristic polynomial of $\psi_\ast$.

One also has a pseudo-Anosov element giving any symplectic matrix, so one has the existence of such an element since $Mod(\Sigma)\to Sp(H_1(\Sigma))$ is surjective.

If you take a mapping class $\psi:\Sigma\to \Sigma$ whose characteristic polynomial of its action on homology $\psi_\ast: H_1(\Sigma;\mathbb{Z}) \to H_1(\Sigma;\mathbb{Z})$ is irreducible, then it cannot extend over a handlebody. The point is that if $\psi= \Psi_{| \Sigma}$, where $\Psi: H\to H, \partial{H}=\Sigma$, $H$ a handlebody, then $\psi$ preserves $K=ker\{H_1(\Sigma)\to H_1(H)\}$, which is an integral Lagrangian subspace of $H_1(\Sigma)$. So the characteristic polynomial of $\psi_{\ast | K}$ divides the characteristic polynomial of $\psi_\ast$. In fact, this argument shows that $\psi$ does not extend over any manifold $M$ with $\partial M=\Sigma$.

Since $Mod(\Sigma)\to Sp(H_1(\Sigma))$ is surjective, one also has a pseudo-Anosov element giving any symplectic matrix (see Sam Nead's comment for one possible argument, or one may restrict to a matrix satisfying Casson's criterion for irreducibility as in Rivin's answer), so there is a pA element satisfying the irreducibility criterion.