Here is an extremely inefficient implementation of Edmonds-Karp (breadth-first search) given as a tropical rational map. (Hopefully I've understood the definition of tropical rational map correctly.)
First, observe that in the usual implementation of Edmonds-Karp, no path is used more than once, since if $p$ is a shortest path with nonzero capacity, the residual edges of $p$ can only be used in a longer path. So if we perform Ford-Fulkerson by iterating over all paths in increasing order of length, the algorithm will terminate with the maximum flow.
Let $\bar{D}$ be the graph obtained by adding, for every edge $a$, a residual edge $\bar{a}$ with target and source reversed. We denote by $\bar{A}$ the set of residual edges. If $a \in \bar{A}$ is a residual edge, we will likewise denote by $\bar{a} \in A$ the edge in the original graph corresponding to $a$. Let $\{p_i\}_{i = 1}^N$ be any fixed enumeration of the acyclic paths from $s$ to $t$ in $\bar{D}$ in increasing order of length.
We will construct our tropical rational map $\mathbb{N}^A \to \mathbb{N}^A$ in three pieces:
First, let $g: \mathbb{N}^A \to \mathbb{N}^{A \sqcup \bar{A}}$ be the augmentation map, taking $c: A \to \mathbb{N}$ to $\bar{c}: A \sqcup \bar{A} \to \mathbb{N}$, where $\bar{c}$ is equal to $c$ on $A$ and identically zero on $\bar{A}$.
For any path $p = (a_1, a_2, \ldots, a_n)$, let $f_p: \mathbb{N}^{A \sqcup \bar{A}} \to \mathbb{N}^{A \sqcup \bar{A}}$ be defined by
$$f_p(c)(a) = \begin{cases}
c(a) - \min(c(a_1), \ldots, c(a_n)) \textrm{ if } a \in p,\\
c(a) + \min(c(a_1), \ldots, c(a_n)) \textrm{ if } \bar{a} \in p,\\
c(a) \textrm{ otherwise.}\end{cases}$$
Note that $f_p(c) = c$ if $c$ is zero on any edge in $p$.
Finally, let $r: \mathbb{N}^{A \sqcup \bar{A}} \to \mathbb{N}^A$ be defined by $r(c)(a) = c(\bar{a})$.
Then the composite $r \circ f_{p_N} \circ \cdots \circ f_{p_1} \circ g: \mathbb{N}^A \to \mathbb{N}^A$ is a tropical rational map sending a capacity function $c$ to a maximum-value $c$-flow.
EDIT:
For completeness, let me give the full argument that if $p$ is a shortest path with nonzero flow, its residual edges can only be used in longer paths. (This is a standard argument in the analysis of the termination and complexity of Edmonds-Karp.)
Denote by $c_i$ the capacity function on $\bar{D}$ after $i$ steps of Edmonds-Karp, and for any vertex $v \in V$, denote by $\ell_i(v)$ the length of the shortest path from $s$ to $v$ in $\bar{D}$ with nonzero $c_i$-flow (i.e., every edge $a$ in the path has $c_i(a) > 0$).
The $(i + 1)$st step of Edmonds-Karp involves flowing along a shortest nonzero-$c_i$-flow path $p$. If $p$ visits vertices $s = v_0, v_1, v_2, \ldots, v_n = t$, then since $p$ is a shortest path, $\ell_i(v_k) = k$ for all $k$. In Edmonds-Karp, we modify the flow capacity by reducing capacity along the edges of $p$ and increasing capacity along the residual edges of $p$. Each residual edge goes from an vertex $v_k$ with $\ell_i(v_k) = k$ to a vertex $v_{k - 1}$ with $\ell_i(v_{k - 1}) = k - 1$.
Now suppose $q$ is a shortest nonzero-$c_{i + 1}$-flow path from $s$ to $t$. Then for each edge from $w_k$ to $w_{k + 1}$ in $q$, $\ell_i(w_{k + 1}) \leq \ell_i(w_k) + 1$, so the length of $q$ is at least $n$ (i.e., the length of $p$), and moreover, if $q$ contains any residual edges of $p$, it must be even longer, since the inequality is strict for those edges. Indeed, by induction, as long as we are flowing along paths of length $n$, we must only be using edges from $w_k$ to $w_{k + 1}$ such that $\ell_i(w_k) = k$ and $\ell_i(w_{k + 1}) = k + 1$.
What this analysis shows is that if we flow along a path $p$ of length $n$ in Edmonds-Karp, we cannot create shorter paths with nonzero flow. Moreover, any other shortest path of length $n$ will not include any residual edges of $p$, so the order in which we try to flow through paths of length $n$ does not matter.
Best Answer
If you are requiring the flow to be a max-flow, this approach is valid; the max-flow at the min-cost of the transformed network must correspond to the max-flow at the max-cost of the original network. So you can just run a min-cost algorithm on the transformed network with the required flow being the max-flow.
The only question is if the min-cost algorithm you want to use can handle negative edge costs. For what it's work, the Wikipedia article on Minimum-Cost Flow states that "most minimum-cost flow algorithms supporting edges with negative costs."
But for this approach to be valid, you must require the max-cost flow to be a max-flow, because it is not true that max-cost flows will always be max-flows. Imagine a network where it is more costly for flow to travel along some cycle, which blocks any flow from getting from the source to the sink. So then calling min-cost max-flow on the transformed network will not give the correct answer.