Well, the diameter of the wormhole's throat should be bigger than the diameter of the planet, for a start. As for tidal forces, here's a rough analysis :
For a Morris-Thorne wormhole, with metric
\begin{equation}
ds^2 = -e^{2\phi(r)} dt^2 + \frac{dr^2}{1-\frac{b(r)}{r}}dr^2 + r^2 d\Omega^2
\end{equation}
Let's take a radially infalling planet, with 4-velocity $V_t = \gamma$, $V_r = \gamma \beta$. The tidal forces for radial motion is (as derived in Visser) :
\begin{equation}
(\Delta a)_\parallel = [(1-\frac{b}{r} (-\phi'' - (\phi')^2) + \frac{1}{2r^2}(b'r-b) \phi'](\Delta \xi)_\parallel
\end{equation}
\begin{equation}
(\Delta a)_\perp = \frac{\gamma^2}{r^2} [(r-b) \phi' + \frac{\beta^2}{2} (b' - \frac{b}{r})](\Delta \xi)_\perp
\end{equation}
For two points separated by $\Delta \xi$.
The metric on a planet will be some static spherically symmetric metric, of the form
\begin{equation}
ds^2 = -A(r) dt^2 + B(r)dr^2 + r^2 d\Omega^2
\end{equation}
For a radial distribution of mass, we have
\begin{equation}
B(r) = (1 - \frac{2m(r)}{r})^{-1}
\end{equation}
And, let's say,
\begin{equation}
A(r) = 1 - \frac{2m(r)}{r}
\end{equation}
to preserve continuity at the boundary ($A$ is somewhat arbitrary depending on initial conditions of the metric).
Hopefully points are not moving on the planet, so the 4-velocity of those points will be $(1,0,0,0)$, and so the radial relative acceleration will be
\begin{equation}
(\Delta a)_\parallel = -(R_{abcd} n^a V^b n^c V^d) (\Delta \xi)_\parallel = -R_{rtrt} (\Delta \xi)_\parallel = -\frac{1}{r^4} ((r-2m(r)) (2m(r) + r(-2m'(r) + rm''(r)))) (\Delta \xi)_\parallel
\end{equation}
The Roche limit for planets not tearing themselves apart is that the tidal acceleration of outside influences should not exceed that of the planet itself. In our case, this means roughly :
\begin{equation}
[(1-\frac{b}{r} (-\phi'' - (\phi')^2) + \frac{1}{2r^2}(b'r-b) \phi'] < -\frac{1}{r'^4} ((r'-2m(r')) (2m(r) + r'(-2m'(r') + r'm''(r'))))
\end{equation}
\begin{equation}
\frac{\gamma^2}{r^2} [(r-b) \phi' + \frac{\beta^2}{2} (b' - \frac{b}{r})]< -\frac{1}{r'^4} ((r'-2m(r')) (2m(r) + r'(-2m'(r') + r'm''(r'))))
\end{equation}
That is a pretty rough analysis, couldn't find too many details on the relativistic Roche limit (in particular this is for perfectly rigid planets, the fluid Roche limit is a bit more forgiving), but that should give you an idea. $r$ and $r'$ are not the same coordinates, in the coordinates of the wormhole moving along $r'$ is roughly as moving away from $r$ in any direction. Of course, this only applies to the Morris Thorne wormhole. Tidal forces in wormholes are linked to their curvatures, and you may recall that a polyhedral wormhole has $R_{abcd} = 0$ everywhere outside of its sides, in which case a cubical wormhole would have to be of side $a = R$ to let a planet pass through.
