Not at all. Consider any binary tree that isn't totally ordered. It's a partial order (so a preorder), but has incomparable elements that will thus have no common upper bound.
Of course, in this particular counterexample, the inverse relation will direct the set, but for a general preorder, neither the given relation nor its inverse need direct the set. For example (thanks, Brian!), consider the set $X=\bigl(\Bbb Z\times\{0\}\bigr)\cup\bigl(\Bbb Z\times\{1\}\bigr)$, with the relation $\precsim$ given by $\langle m,j\rangle\precsim\langle n,k\rangle$ iff $j=k$ and $m\leq n$ (where $\leq$ is the standard non-strict order on $\Bbb Z$). Both $\precsim$ and $\precsim^{-1}$ preorder $X$ (isomorphically), but do not direct the set.