It seems the answer is known, so I will just provide a reference and some comments. A google search with the terms Fenn's table theorem rectangle brings up a paper by:
Bill Baritompa, Rainer Löwen, Burkard Polster, and Marty Ross, titled
Mathematical Table Turning Revisited,
From their introduction: We prove that given
any rectangle, any continuous ground and any point on the ground, the rectangle can be positioned such that all its vertices are on the ground and its center is on the vertical through the distinguished point. This is a mathematical existence result and does not provide a practical way of actually finding a balancing position.
Towards a proof, they define "a mathematical table" using a rectangle, as follows.
In the mathematical analysis of the problem, we will first assume that the ground is the graph of a function $g : \Bbb R^2 \to \Bbb R$, and that a mathematical table consists of the four vertices of a rectangle of diameter $2$ whose center is on the $z$-axis.
They state further down: This result is a seemingly undocumented corollary of a theorem by Livesay , which can be phrased as follows: For any continuous function $f$ defined on the unit sphere, we can position a given mathematical table with all its vertices on the sphere such that $f$ takes on the same value
at all four vertices.
 Livesay, George R. On a theorem of F. J. Dyson. Ann. of Math. 59 (1954), 227–229.
The following is Theorem 3 from Livesay 
(where $S_2=S^2$ is the sphere, and $E_1=\Bbb R$ is the real line).
Let $f:S_2\to E_1$ be continuous, $0 < \theta \le 90^\circ$.
Then there exist two diameters of $S_2$ subtending the angle $\theta$,
such that the four end points $y_1, y_2, y_3, y_4$ of these diameters satisfy
$f(y_i) = f(y_j), i, j = 1,...,4$.
Going back to the paper by Baritompa, Löwen, Polster, and Ross, they use
Livesay's Theorem as follows. Given any continuous ground function
$g : \Bbb R^2 \to \Bbb R$ and any "mathematical table" (a rectangle with diagonals of length $2$) they construct $f:S^2\to\Bbb R$ by
$f(x,y,z)=z-g(x,y)$. Then apply Livesay's Theorem to position the
given mathematical table with all its vertices on the sphere and
such that $f$ takes on the same value at all four vertices.
The picture (as I interpret it) is that the table is the given rectangle
(assuming without loss of generality that the diagonals have length $2$),
and the above proof positions the vertices of the table on the unit sphere in such a way that if we assume that the legs are vertical, then the table is perfectly balanced on the given ground $g$. This was confusing a bit initially, since I am used to think that the legs ought to be perpendicular to the table (and the table need not be horizontal, so the legs would not need to be vertical). But this is not really a problem (the authors don't seem to comment on it, but it appears to be a triviality ... and actually they do comment later in their paper about mathematical vs real tables), once we balance the table (not necessarily horizontal) so that the legs a vertical, then we could "move" the table so that the bottom of each leg remains fixed on the ground, while the table and the top of each leg move until the legs become perpendicular to the table. (I was a bit lost for a while to see exactly what was proved in which paper and how the result follows, so now that I think I figured it, I will include my interpretation here, for convenience.)
I indicated in a comment that this subject also related to the inscribed square problem, so I include a few more links (about both the squarepeg problem, and about versions of the table theorem), in case someone might find them useful.
A Survey on the Square Peg Problem, by Benjamin Matschke,
Notices of the AMS Volume 61, Number 4, p.346-352.
In particular note Conjecture 13 (Table problem on $S^2$) there:
Suppose $x_1,x_2,x_3,x_4\in S^2\subset\Bbb R^3$ are the vertices of a square
that is inscribed in the standard $2$-sphere, and let
$h : S^2\to\Bbb R$ be a smooth function.
Then there exists a rotation $\rho\in SO(3)$
such that $h(\rho(x_1))=h(\rho(x_2))=h(\rho(x_3))=h(\rho(x_4))$.
So far this result has been proven only when
$x_1,x_2,x_3,x_4$ lie on a great circle (see Dyson).
Balancing acts, by Mark Meyerson,
Topology proceedings, Volume 6, 1981, pages 59-75
Remarks on Fenn's "the table theorem" and Zaks' "the chair theorem",
by Mark D. Meyerson,
Pacific Journal of Mathematics, Volume 110, Number 1 (1984), 167-169,
The Table Theorem, by Roger Fenn,
Bull. London Math. Soc. (1970) 2 (1): 73-76, doi: 10.1112/blms/2.1.73
Dyson, F. J. Continuous functions defined on spheres.
Ann. of Math. 54 (1951), 534–536.