No, if $M^\circ$ is orientable then $M$ is orientable, and hence $\partial M$ is orientable. Here's two proofs. They're both secretly the same.
1) The obstruction to orientability is the first Stiefel-Whitney class $w_1(M) := w_1(TM) \in H^1(M;\mathbb Z/2)$. Fact: if $f$ is a smooth map, then $f^*w_1(\xi) = w_1(f^*\xi)$. Then $TM^\circ = \iota^* TM$, where $\iota$ is the inclusion map; and note that $\iota$ induces an isomorphism on cohomology, as it's a homotopy equivalence. So under the obvious identifications of cohomology rings, $w_1(TM^\circ) = w_1(TM)$, so $TM^\circ$ is orientable iff $TM$ is.
2) An orientation is a nonvanishing section of the top exterior power of the tangent bundle $\Lambda^n TM$. Given a nonvanishing section of $\Lambda^n TM^\circ$, put a fiberwise metric on $TM$ and suppose this section is of unit norm. Then in a small neighborhood of any point $p \in \partial M$, there is one and only one way of extending this to a section of $TM$. (Remember that locally in a chart $U$, the unit sphere bundle of $\Lambda^n TM$ looks like $\mathbb Z/2 \times U$.) So extend your section! There is no ambiguity, and this is again a smooth section by definition of the fiberwise metric varying smoothly. So one can always extend an orientation of $\Lambda^n TM^\circ$ to $\Lambda^n TM$ if you desire.
You can rewrite 2) in terms of nonvanishing differential forms of top degree, if you like. This may be more intuitive than thinking about it in terms of bundles. The replacement for the metric on $\Lambda^n TM$ is to scale your choice of form so that $\omega_p(X_1, \dots, X_n) = \pm 1$ if the $X_i$ are an orthonormal frame at $p$.