What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms,
$$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$
could indeed be called the 'dipolar component', and it will work in any region where there are no charges other than test charges. You chould note in particular that this object is intimately related to the electric field, as
$$\Delta\phi=-\nabla \mathbf{E}\cdot(\mathbf{r}-\mathbf{r}_\text{c}).$$
Because of this, using the dipolar component will work as long as you evaluate it at the dipole separation.
On the other hand, if you take the vector
$$\mathbf{\Delta\phi}=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})\mathbf{e}_j$$
and then dot that with the atoms' positions, then you will get an interaction energy
$$
U
=\sum_a\sum_j\frac{\partial \phi}{\partial x_j}(x_{a,j}-x_{\text c,j})p_j
=\frac2q\sum_j\frac{\partial \phi}{\partial x_j}p_j^2
$$
which is manifestly wrong.
So, bottom line: no.
You haven't really explained what $Z$ is, so let me know in the comments if my definitions are OK with your problem.
For the ground state of the hydrogen atom, represented by $|nlm\rangle=|100\rangle$ (with a wavefunction $\psi_{100}$), we try to work out the first-order shift in the energy caused by the perturbation of an electric field along the $z$-axis. This field gives a perturbing potential $Z=e\Phi \hat{z}$ (where $e$ is the charge of the electron, $\Phi$ is the value of the field in the $z$ direction, and $\hat{z}$ is the position operator).
The first-order shift in the ground state energy eigenvalue is given by
$$\Delta E_0=\langle 100|Z|100\rangle=\langle100|e\Phi\hat{z}|100\rangle=\int\psi_{100}^*(r)e\Phi\hat{z}\psi_{100}(r)dr^3=e\Phi\int z(r)|\psi_{100}(r)|^2dr^3$$
The last equality above is simply due to: $\hat{z}\psi=z\psi$.
Now, when we say that $\psi_{100}$ has a definite parity, what we mean is that it is either an even function or an odd function. A function with indefinite parity is one that is neither even nor odd. But whatever the parity of $\psi_{100}$ actually is, $|\psi_{100}|^2$ will be even (even$\times$even$=$even, odd$\times$odd$=$even). And of course, $z(r)$ is odd. Overall, the integrand is an odd function (even$\times$odd$=$odd). An an odd function integrated over the whole space is $0$.
Therefore $\Delta E_0=0$, there is no first-order shift in the ground state of hydrogen, i.e. there is no linear Stark effect.
(You could also try to show, by the same sort of reasoning, that there is no linear Stark effect for any atom in a non-degenerate energy eigenstate).
Best Answer
The linearity in hydrogen is due to the fact that you can take energy eigenstates that have net dipole moments. This requires a superposition of states with different values of $\ell$. It is only because of the accidental degeneracy of the Coulomb problem that there are $n=2$ energy levels with different values of $\ell$. The states $|2S\rangle\pm|2P_{z}\rangle$ have dipole moments (or order $ea_{0}$) along the $z$-direction. The expectation values of the electric operator $-e{\cal E}z$ in these states give the first order Stark shifts in hydrogen.
However, in atoms with more than one electron $|2S\rangle$ and $|2P\rangle$ are not degenerate, so it is not possible to form a superposition state that is also an energy eigenstate. (In fact, in hydrogen the states are not degenerate either, because of the combine effects of spin-orbit coupling and the Lamb shift resulting from quantum fluctuations in the electric field.) This means the perturbative energy shifts for small fields must be quadratic in ${\cal E}$ However, when the electric field ${\cal E}$ becomes large enough the energy splittings between the different $\ell$ states become negligible. Then the effects are very similar to what they would be if the different $\ell$ states were truly degenerate. However, this is, strictly speaking, beyond the perturbative regime, and the energies cannot be read off directly from the perturbative formulas (although on can get the limiting linear expressions by using first-order perturbation theory and neglecting the energy spliting between the different $\ell$ states, just as the relativistic and Lamb shift corrections were ignored in the above discussion of hydrogen).