The Gauss--Lucas Theorem states that all zeros of a degree $n$ complex polynomial $p(z)$ are contained in the convex hull of the zeros of $p$. By iteration, this implies that the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$ are contained in the convex hull of the zeros of $p$.
The Riemann--Hurwitz Theorem (among others) implies that if a tract $D$ of $p$ (namely a component of the set $\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) contains all the zeros of $p$ in its bounded face, then all the critical points of $p$ are contained in $D$.
My conjecture is that in fact, if $D$ is a tract of $p$ and contains all the zeros of $p$, then $D$ also contains all the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$.
This certainly does not follow by straight-forward iteration, since in general there need not be a tract of $p'$ containing all the zeros of $p'$ which is contained in $D$. It seems that the tracts and level curves of $p'$ do not interact very nicely with the tracts and level curves of $p$ (even worse for $p'',p''',\ldots$).
I have taken a look at attempting to apply the Cauchy Integral Formula (some sort of integration by parts application perhaps?), but don't seem to be able to make progress there. Any ideas for proof or counter-example?
This is a community wiki answer to remove this question from the unanswered list: Bobby Ocean posted a counter-example at Mathoverflow: Set $p(z) = (z^4+2z^2+2) (z-1)$. Then $D:= \{ z : |p(z)|<1.45\}$ is connected, but some of the zeroes of $p'''$ and $p''''$ are outside $D$.