EDIT: The original question appears to have been worded much too abstractly in terms of constant and coconstant morphisms to admit a precise answer (for details see history), the new question has been restated in a much more concrete form after receiving comments.
Restated question: Can morphisms in a concrete category admit mappings that are not functions?
It arose when dualizing a constant function $f:X\longrightarrow Y$ in $\mathcal C$ into a dual morphism $F(f):F(Y)\longrightarrow F(X)$ in $\mathcal B^{op}$ under a contravariant dually equivalent functor $F:\mathcal C\longrightarrow \mathcal B^{op}$ with $\mathcal C \simeq \mathcal B^{op}$ (dually equivalent). $F(f)$ then should map a constant $c\in Y$ in such a way that $F(f)(c):F(Y)\longrightarrow F(X)$ outputs its entire codomain $F(X)$, which is at the same time both the image and the output of $F(f)(c)$. This appears to be a multifunction rather than a function.
For more concreteness, $\mathcal B^{op}$ can be taken as $Set^{op}$. A construction is given on Wikipedia on how to embed $Set^{op}$ into the category Rel of sets and relations. Both Rel and $Set^{op}$ are known to be concrete categories. Then the morphism $F(f)$ in $Set^{op}$ that is dual to a constant function $f$ in $\mathcal C$ can be a multifunction, because relations need not be functions. Some more specifics and references are given in my comments to this post.
To the most recent version of the question, the answer is that it depends on what you mean by "concrete category". The most common usage of the term means, specifically, a category equipped with a faithful functor to $\mathbf{Set}$, so the answer would be no. If you've been reading Abstract and Concrete categories, what most folks now call "concrete categories" are what those authors refer to as "constructs."
There is also the more general notion of "concrete over $\mathcal D$," meaning it's a category equipped with a faithful functor into $\mathcal{D}$. In this case, sure, you could have a concrete category over $\mathbf{Rel}$, or over the trivial category (as with preorders), or whatever else. For instance, $\mathbf{Set}$ is concrete over $\mathbf{Rel}$ in this sense, or over $\mathbf{Eff}$.