If $X=\mathbb{V}(I)$ is given by the ideal $I$, then the $m$th symbolic power $I^{[m]}$ consists of all those functions vanishing to multiplicity $m$ at the generic point of $X$. Thus a hypersurface $\mathbb{V}(F)$ will be singular along $X$ if and only if $F\in I^{[2]}$. (Thanks to Zach Teitler for pointing this out in the comments below.) To give a counterexample it is therefore enough to find an $X$ for which $I^2\subsetneq I^{[2]}$.

A (non-irreducible, affine) example is $X=\mathbb{V}(xy,yz,zx)$, the union of the three coordinate axes in $\mathbb{A}^3$ and $F=xyz\notin I^2$. Then the hypersurface $\mathbb{V}(xyz)$ has multiplicity $2$ on every component of $X$. In general if $X\subset \mathbb{P}^3$ is an irreducible space curve which is smooth apart from a triple point (i.e. a point locally analytically isomorphic to the example above) then there should also be function $F\in I^{[2]}\setminus I^2$ as above.

(P.S. If you want to compute $I^{[m]}$ explicitly (e.g. in the case you are interested in, of a curve $X$ of small degree) then you can ask your favourite computer algebra package for the primary decomposition of $I^m$. The unique primary component supported on the whole of $X$ will be $I^{[m]}$.)