This is true and actually has nothing to do with the morphism. It's a simple fact about divisors and their associated reflexive sheaves.

So, $\omega_{X/C}$, the reflexive sheaf of rank $1$ associated to the Weil divisor $K_{X/C}$ is the reflexive hull of $\Omega_{X/C}^r$. In particular, there exists a natural morphism
$$
\Omega_{X/C}^r \to \omega_{X/C},
$$
which is neither necessarily injective nor surjective, but can be decomposed as a surjective morphism followed by an injective one.
$$
\Omega_{X/C}^r \twoheadrightarrow (\Omega_{X/C}^r)/({\rm torsion}) \hookrightarrow \omega_{X/C},
$$

It is easy to see that if you have a morphism
$$
\Omega_{X/C}^r \to \mathscr O(D),
$$
to a torsion-free sheaf, then it factors through $(\Omega_{X/C}^r)/({\rm torsion})$. On the other hand, since $\omega_{X/C}$ is the reflexive hull of $\Omega_{X/C}^r$ and hence also of $(\Omega_{X/C}^r)/({\rm torsion})$, any reflexive sheaf (e.g., a line bundle) that contains $(\Omega_{X/C}^r)/({\rm torsion})$, also contains $\omega_{X/C}$, which translated to divisors means that $K_{X/C}\leq D$.

This means that what you would like is actually true. In fact, $D$ doesn't even have to be Cartier, it works if it is Weil divisor.

**Appendix** (in response to the request in the comments):

As I mentioned, the morphism $\Omega_{X/C}^r \to \omega_{X/C}$ is simply the double dual. For any sheaf $\mathscr F$ you have a natural map
$\mathscr F\to \mathscr F^{\vee\vee}$. This is it.

Another way to think about it is to consider the embedding of the open set $\imath: U=X\setminus {\rm Sing}\, X\hookrightarrow X$ and then
$\mathscr F=\imath_*\imath^*\mathscr F$ so the above map is the natural map given by $1\to \imath_*\imath^*$.

For more on these issues, look at these MO answers:
S2 property and canonical sheaf.