Let $M,N$ be diffeomorphic compact Riemannian manifolds, and let $f:M \to N$ be a nonexpanding map (i.e Lipschitz with constant $1$). Assume that
$(1)$ $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $d(f(p),f(q)) < d(p,q)$.
$(2)$ The image $f(M)$ is a submanifold of $N$. (Note I do not assume $f$ is smooth).
Is it true that $\operatorname{Vol}(f(M))<\operatorname{Vol}(M)$?
If it helps, we can assume for start $M,N$ have empty boundary.
Note that if we do not assume $M,N$ are diffeomorphic then the answer is negative:
$f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$ is strictly nonexpanding but $\operatorname{Vol}(f([0,2\pi])=\operatorname{Vol}(\mathbb{S}^1)=\operatorname{Vol}([0,2\pi])$.
Partial result:
$(1)$ In the case where $M=N$ (as Riemannian manifolds), the answer is positive.
Assume otherwise; Then $\operatorname{Vol}(f(M))=\operatorname{Vol}(M)=\operatorname{Vol}(N)$, hence $f(M)=N$, i.e $f$ is surjective. (Otherwise $f(M)$ will be a closed subset of $N$, strictly contained in $N$, contradicting the equality of volumes).
So, $f$ is a surjective nonexpanding map from a compact metric space to itself, thus an isometry. (See Burago-Burago-Ivanov's "A course in metric geometry", theorem 1.6.15).
$(2)$ In the case the manifolds are one-dimensional, and $f$ is surjective, the answer is positive:
Assume otherwise. Then $\operatorname{Vol}(N)=\operatorname{Vol}(M)$. Since every two compact connected one-dimensional Riemannian manifolds with equal volumes are isometric, there exists an isometry $\phi:N \to M$. Thus, $f \circ \phi:N \to N$ is a surjective nonexpanding map from a compact metric space to itself, hence an isometry.
Result $(1)$ suggests it might be easier to handle the case where $\operatorname{Vol}(M)=\operatorname{Vol}(N)$. The question then becomes equivalent to the following one:
Can a strictly nonexpanding map between two compact Riemannian manifolds of the same volume be surjective?
Here is @AntonMalyshev counterexample in detail (for the case of manifolds with corners):
Let $M$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,1/3]$. This is a manifold with corners (for example, a small enough neighborhood of the point $(0,1/3)\sim (1,1/3)$ is diffeomorphic to $[0,\infty)\times[0,\infty)$). The distance between $(0,1)$ and $(1,1)$ can easily be verified to be $1$.
Let $N$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,2/3]$. Here the distance between $(0,1)$ and $(1,1)$ is at most $2/3$, as shown by the curve $$ \gamma(t) = \begin{cases} (0,1)(1-t) + (0,2/3)t & t\in[0,1] \\ (1,2/3)(2-t) + (1,1)(t-1) & t\in[1,2]. \end{cases} $$
Obviously $M$ and $N$ are diffeomorphic as manifolds with corners, and have the same volume. The trivial map $M\to N$ is surjective, volume preserving and strictly non-expending according to your definition.