No = there are no non compact smooth manifolds which have the property that every Riemannian metric is geodesically convex.

Note that any non compact $n$-manifold $M$
contains a closed subset homeomorphic to $[0,\infty)\times D^{n-1}$
Indeed choose a bounded metric on the manifold $M$
and a minimizing geodesic $\gamma\colon[0,\ell)\to M$ and then pass to the appropriate closed neighborhood of $\gamma$ [see (*)]

Now it is easy to construct a metric $g$ on $[0,\infty)\times D^{n-1}$ which is not geodesically convex [see (**)].
One can also ensure that there is a pair of points say $x,y\in[0,\infty)\times D^{n-1}$
without a $g$-geodesic between such that intrinsic distance from $x$ to $y$ in $ ([0,\infty)\times D^{n-1},g)$ is much smaller than distance from $x$ to $\partial \left([0,\infty)\times D^{n-1}\right)$.
The later condition implies that no matter how you extend this metric to whole $M$, it will fail to be geodesically convex at the pair $(x,y)$.

(*) Given a function $r\colon[0,\ell)\to\mathbb R_{>0}$ consider the subset $W_r$ in the normal bundle to $\gamma$ formed by all the normal vectors $v$ at $\gamma(t)$ such that $|v|<r(t)$. Note that $W_r$ is homeomorphic to $[0,\infty)\times D^{n-1}$ and for small enough functions $r$, the restriction of normal exponential map to $W_r$ is an embedding.

(**) Take the induced metric for $h\colon [0,\infty)\times D^{n-1}\to \mathbb R\times \mathbb R^{n-1}$ where $h(t,x)= (f(t,x),x)$ and $f(x,t)$ is an increasing in $t$ and $s(x)=\lim_{t\to\infty} f(x,t)$ is very small near the center of the disc.