Is it true that on any complete, non-compact Riemannian manifold $(M, g)$ there is a smooth vector field X such that $\operatorname{div} X=1$?
My definition of the divergence is the coordinate one (that turns out not to depend on the choice of coordinates, of course), and that does not require M to be orientable.
If you assume that $M$ is connected (or at least that $M$ has no compact components), it's true.
Let's assume first that $M$ is orientable. Let $\mu_g$ denote its Riemannian volume form. Because the top-degree de Rham cohomology on a connected, noncompact smooth manifold is trivial (see, e.g., my Introduction to Smooth Manifolds, 2nd ed., Theorem 17.32), there is a smooth $(n-1)$-form $\eta$ such that $d\eta=\mu_g$. Set $X = (-1)^{n-1}(\mathop{*}\eta)^\sharp$ (where $\ast$ is the Hodge star operator and $\sharp$ is the index-raising operator determined by the metric). It follows that $$ \operatorname{div} X = \mathop{*} d \ast X^\flat = \mathop{*} d \big((-1)^{n-1}\ast\mathop{*} \eta\big) = \mathop{*} d\eta = \mathop{*} \mu_g = 1. $$
Now if $M$ is nonorientable, let $\widehat \pi \colon \widehat M\to M$ be the oriented double cover of $M$, and let $\widehat g=\widehat\pi^*g$. Then by the previous argument, there is a vector field $\widehat X$ on $\widehat M$ satisfying $\operatorname{div} \widehat X = 1$. Let $\alpha\colon \widehat M\to \widehat M$ denote the nontrivial covering automorphism of $\widehat M$, and let $\widehat X_1 = \tfrac12 (\widehat X + \alpha_* \widehat X)$. Because $\alpha$ is an isometry of $\widehat g$, it follows that $\operatorname{div}\widehat X_1=1$. Moreover, since $\widehat X$ is invariant under $\alpha$, it can be "pushed down" to a well-defined vector field $X$ on $M$, which satisfies $\operatorname{div} X=1$ because $\widehat \pi$ is a local isometry from $(\widehat M,\widehat g)$ to $(M,g)$.
If $M$ has a compact component, a vector field $X$ satisfying $\operatorname{div} X = 1$ would contradict the divergence theorem, which would require $\int_M (\operatorname{div} X)\, \mu_g = 0$.