Let $G$ be a finite non-abelian simple group. Question: 



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This seems to be the case for alternating groups (take $M$ a stabilizer of a $4$-set) and sporadic simple groups (look at tables in ATLAS). For groups of Lie type in characteristic $2$ it seems to me that one could choose $M$ to be a maximal parabolic subgroup.


Also if the answer to the question is no, what about the weaker question of $M$ containing a non-trivial normal abelian subgroup? And if the answer to either question is yes, is there a proof without classification?



Maximal subgroups of simple groups with normal $2$-subgroups
Unfortunately I made a mistake in my original answer. The answer to the question is no, but the only finite simple groups that do not have a maximal subgroup with nontrivial normal 2-subgroup are the groups $L\_p(3)$ with $p$ a prime and $p \ge 5$.


Let's go through the finite nonabelian simple groups.


As you say, for $G=A\_n$ we can take $M$ to be the stabilizer of a set of size $4$.


You have checked the sporadic groups in the ATLAS.


If $G$ is of Lie type in characteristic $2$, then we can take $M$ to be a maximal parabolic subgroup.


So we can assume from now on that $G$ is of Lie type in odd characteristic.


The geometric-type maximal subgroups of the classical groups are listed in the book by Kleidman and Liebeck. Some of these are not maximal in small dimensions. The complete classification in dimensions up to $12$ was completed in a book by myself, Bray and Roney-Dougal. So we have enough information to check the result in groups of odd characteristic, and in fact imprimitive groups can be used for $M$ in most cases.


In some of the lists of $M$ below, there are a few small $n$ and $q$ for which $M$ is not maximal, but in each of these cases you can check that there is some other maximal subgroup that works - more details on request.


For $G=L\_2(q)$ we have a dihedral subgroup of order $(q-1)$ or $(q+1)$.


Form now on, the subgroup $M$ is described as a subgroup of the relevant quasisimple matrix group, so we have to take it modulo scalars to get the corresponding maximal of the simple group.


In general, for $G=L\_n(q)$ with $n \ge 3$ and $q \ge 5$, we can take $M$ to be the imprimitive group with structure $(q-1)^{n-1}:S\_n$.


For $L\_n(3)$, the subgroup with this structure is not maximal (it preserves an orthogonal form). If $n=rs$ is composite with $r \ge 2$ and $s>2$, then there is an imprimitive subgroup ${\rm GL}\_r(3) \wr S\_s$ of ${\rm GL}\_n(3)$, and its intersection with ${\rm SL}\_n(3)$ projects onto a maximal subgroup of $L\_n(3)$ with a nontrivial normal $2$-subgroup.


Also $L\_3(3)$ has $S\_4$ as maximal subgroup and $L\_4(3)$ has $S\_4 \times S\_4$, but for primes $p \ge 5$, $L\_p(3)$ has no maximal with the required property. This can be seen from the complete list of its maximal subgroups, and I checked it directly by computer for $p=5,7,11$.


$G=U\_n(q)$, we can take $M$ to be the imprimitive subgroupn with structure $(q+1)^{n-1}:S\_n$, and this really is maximal for all $q \ge 3$.


$G=S\_{2n}(q)$, take $M={\rm Sp}\_2(q)^n:S\_n$.


The orthogonal groups all have reducible maximal subgroups that are stabilizers of non-degenerate subspaces of dimension $2$ with nontrivial normal $2$-subgroups.


I know very little about the maximal subgroups of the exceptional groups of Lie type, but I looked at Table 5.2 of 


M.W. Liebeck, J. Saxl and G.M. Seitz, “Subgroups of maximal rank in finite
exceptional groups of Lie type”,
Proc. London Math. Soc.
65
(1992), 297-325.


In all cases there appears to be a maximal subgroup that looks like an imprimitive maximal of a classical groups, and has normal subgroups with structure $(q-1)^n$ or $(q+1)^n$, where $n$ is the Lie rank of the group.


