This is a great question! (Or rather, I quite liked it. Reasonable people can differ on this, I suppose.) In fact, local considerations (or even semi-local considerations like understanding the symplectic geometry of a Weinstein neighbourhood of the hypersurface in question) can't suffice to rule out the possibility of the existence of a transverse symplectic vector field, as the example of separating and non-separating circles in the torus shows; any two simple closed curves in the torus have neighbourhoods which are symplectomorphic, since they're Lagrangian, but a meridian of the torus obviously admits a transverse symplectic vector field induced by translation in the universal cover in the direction of the equator, while a separating loop admits no transverse symplectic vector field by area considerations.

So global considerations are in order, and with that in mind, we can give a nice account of when a hypersurface admits a transverse symplectic vector field. Recall that any hypersurface $\iota: N \rightarrow M$ in a symplectic manifold $(M,\omega)$ is coisotropic and so carries a one-dimensional line bundle $L \rightarrow N$ whose fiber at $p$ is $(T_pN)^\omega$. In this setting, we have:

**Proposition:**
There exists a symplectic vector field $X \in \mathcal{X}(M)$ which is transverse to $\iota(N)$ if and only if there exists a closed one-form $\sigma \in \Omega^1(N)$ such that:

(1) $\sigma$ is non-vanishing on non-zero vectors in $L$ (so in particular, $\sigma$ is non-exact), and

(2) $[\sigma]$ lies in the image of the induced map on cohomology. ie: $[\sigma] \in im \big( \iota^*: H^1(M;R) \rightarrow H^1(N;R) \big)$

The details are easy enough to check and relies simply on the observation that $X \in \mathcal{X}(M)$ is a symplectic vector field, transverse to $N$ if and only if the one-form $\tilde{\sigma}:= X\lrcorner\omega$ is closed and non-vanishing on non-zero vectors in $L$ (if $Y$ is any non-vanishing section of $L$, then $X(p)$ lies outside $<Y(p)>^\omega=T_pN$), along with the fact that the flux map which sends symplectic isotopies to the average of their generating 1-forms is a surjection onto $H^1(M;R)$.

As to the preponderance of such objects, in some sense they aren't that uncommon; any odd-dimensional manifold $N$ which has a closed $1-$form which doesn't vanish on some one-dimensional distribution on $N$ embeds in a symplectic manifold which has a symplectic vector field transverse to $N$, because we can put a symplectic structure on $N \times S^1$ such that the vector field $\partial_t$ is symplectic and transverse to $N \times \lbrace 0 \rbrace$. But I'll admit that I'm curious to know what more can be said about this aspect of the question.