No. Define a Riemannian metric tensor $g$ on $\mathbb R^2=\{(x,y)\}$ by
$$
g(x,y) = \begin{pmatrix} 1 & 0 \\ 0 & f^2(x) \end{pmatrix}
$$
where $f:\mathbb R\to\mathbb R$ is a positive smooth even function such that
$f(x) = \cos x$ for $|x|\le 1$ and $f''(x)/f(x)$ increases after $x=1$. Let $N$ be an open segment of length $\pi$ in the $y$-axis, e.g. the one between points $A=(0,0)$ and $B=(0,\pi)$.

(The plane with this metric is isometric to the universal cover of a surface of revolution that looks like a unit sphere with two infinite tubes attached near a pair of opposite points. Note that the Gaussian curvature $K$ is given by $K=-f''/f$, so $K\le 1$ everywhere.)

The strip $\{|x|\le 1\}$ is isometric to the universal cover of a neighbourhood of the equator of the standard sphere, so there are plenty of geodesics between $A$ and $B$. On the other hand, using Clairot integral and the fact that the Gaussian curvature does not exceed 1, it is easy to see that no geodesic can intersect the $y$-axis at two points with distance less than $\pi$ between them. Hence $N$ is totally convex but its closure is not.