Yes, though I'm not sure if you'll like this example. Take any incomplete normed space and equip it with the trivial order, where $x \le y$ iff $x=y$ (i.e. any distinct elements are incomparable). This satisfies the axioms of an ordered vector space, and $X^+ = \{0\}$ which is of course complete.
For a somewhat natural example of this, let $X$ be the vector space of sequences of real numbers which are eventually zero and which sum to zero. That is, its elements are of the form $(x_1, \dots, x_n, 0,0, \dots)$ with $\sum_{i=1}^n x_i = 0$. This is clearly a vector space, and when equipped with the natural pointwise ordering (i.e. $x \ge y$ if $x_i \ge y_i$ for all $i$) the only positive element is 0, so the ordering is trivial. $X$ has countable dimension (a Hamel basis is given by sequences of the form $(0,\dots, 0,1,-1,0,\dots)$) and so by the Baire category theorem $X$ is incomplete in any norm. (You could use an $\ell^p$ norm, for instance.)