Not necessarily: consider a 3-D space, let $L$ be the $x$ axis, and let the convex set be $$X=\bigl\{(x,y,z):z\geqslant (\max\{0,y + e^x\})^2\bigr\}.$$ This is the region above the graph of a convex function; here's a 3-D plot. After projecting on the $yz$ plane, the convex set becomes $$X' = \{(y, z) : y < 0 , \, z \geqslant 0\} \cup \{(y, z) : y \geqslant 0, z > y^2\}.$$ There is no line in the $yz$ plane that contains the origin and does not intersect $X'$, and hence there is no hyperplane that contains the $x$ axis and does not intersect $X$.