Suppose the derivative of a functional is given by
\begin{equation*} \int_{\Omega}(\vec{v}.\nabla u)|\nabla u|^{p-2} \phi=\int_{\Omega}\nabla.(u\vec{v})|\nabla u|^{p-2} \phi,~\phi\in W_0^{1,p}(\Omega) \end{equation*}
where the vector field $\vec{v}$ (which is known) is irrotational, i.e., $\nabla.\vec{v}=0$, then what is the functional?. The derivative is computed at $u$ and the argument of the derivative is $\phi$.
Call your putative variational derivative $E[u] = (\vec{v}\cdot \nabla u) |\nabla u|^{p-2}$. To be an actual variational derivative of a local functional it is a necessary and sufficient condition (these are the Helmholtz conditions, and it is at least easy to check their necessity) for $E[u]$ to have a (formally) self-adjoint linearization $J[\delta u] = \delta E[u]$. Unless I calculated incorrectly, in this case $J^*[\delta u] \ne J[\delta u]$, so your $E[u]$ cannot be the variational derivative of any local functional.
