The simplest example is the following. Take $X = A^3$ with coordinates $(x,y,z)$, and let $E = Ker(O_X \oplus O_X \oplus O_X \stackrel{(x,y,z)}\to O_X)$. Let $U$ be the complement of the point $(0,0,0) \in X$. Then $E_{|U}$ is a vector bundle. On the other hand, $E$ is not a vector bundle, but $E^{**} \cong E$, hence $E$ is the reflexive envelope of $i_*i^*E$, and thus there is no vector bundle on $X$ extending $E_{|U}$.

[Edit by Anton: I just spent some time digesting some pieces of the above answer, so figured I'd include the results for future readers similar to me.]

**("$E$ is not a vector bundle")** The sequence $O_X\xrightarrow{\pmatrix{z\\ y \\ x}}O_X^3\xrightarrow{\pmatrix{y & -z & 0\\ -x & 0 & z\\ 0 &x&-y}}O_X^3\xrightarrow{\pmatrix{x& y& z}}O_X$ is exact, so $E$ is the cokernel of the first map. Since taking fibers commutes with taking cokernels, we compute that $E$ has 2-dimensional fibers away from the origin, and 3-dimensional fiber at the origin.

**("$E^{**}\cong E$")** Note that $E$ is $S_2$ (i.e. sections defined away from codimension 2 extend uniquely) since it is the kernel of a map from an $S_2$ sheaf to a torsion-free sheaf (the section of $O_X^3$ extends uniquely, and its image is zero away from codimension 2, so must be zero, so the extended section is in $E$). Note also that the dual of any sheaf is $S_2$ (if $\phi\colon F\to O_X$ is defined on an open set $V$ with codimension 2 complement and $s$ is a section, $\phi(s)$ must be the unique extension of $\phi(s|_V)$ as a section of $O_X$), so $E^{**}$ is $S_2$. The canonical map $E\to E^{**}$ is then a map of $S_2$ sheaves which is an isomorphism away from codimension 2, so it must be an isomorphism.

**("and thus there is no vector bundle on $X$ extending $E|_U$")** If $F$ is an $S_2$ extension of $E|_U$ (i.e. $i^*F=i^*E$), then there is a map $F\to i_*i^*E\to (i_*i^*E)^{**}=E$ which is an isomorphism over $U$, so is an isomorphism by the argument in the previous paragraph. A vector bundle extension would be a different $S_2$ extension.