Here's another look at the case $m=2$. Maybe it can be used to shed more light on Torsten's nice example. $S^2E$ is part of an exact sequence $0\to \Lambda^2E\to E\otimes E\to S^2E\to 0$. $\Gamma^2E$ is part of an exact sequence $0\to \Gamma^2E\to E\otimes E\to \Lambda^2E\to 0$. The composition of $E\otimes E\to \Lambda^2E \to E\otimes E$ is $1-T$ where $T$ is the involution $x\otimes y\mapsto y\otimes x$. Its kernel $\Gamma^2E$ may be considered as the symmetric bilinear forms on $E^*$, while its cokernel $S^2E$ is the quadratic forms. If $2=0$ then the equation $(1-T)(1+T)=0$ says that the image of $1-T$ is contained in the kernel of $1-T$; we have $\Lambda^2E$ injecting into $\Gamma^2E$. We have in fact an exact sequence $0\to \Lambda^2E\to\Gamma^2E\to E'\to 0$, and another one $0\to E'\to S^2E\to \Lambda^2E\to 0$, where I am writing $E'$ for "$E$ twisted by Frobenius".
If you're not in characteristic $2$ then there's no reason for $S^2E$ (or $\Gamma^2E$) to have a proper nontrivial subbundle.
Added: In the case when $E$ is the tangent bundle of $P^2$, or alternatively the rank $2$ quotient bundle of a trivial rank $3$ bundle which, as Torsten mentions, is the tangent bundle twisted by a line bundle, I believe it is not hard to work out by hand that the only global maps $E\otimes E\to E\otimes E$ are the linear combinations of the identity and the involution $v\otimes w\mapsto w\otimes v$. Of these, the only ones that kill the image of $\Lambda^2 E$ and so give a map $S^2E=coker(\Lambda^2E\to E\otimes E)\to E\otimes E$ are the multiples of $v\otimes w\mapsto v\otimes w+w\otimes v$, so that the only maps $S^2E\to \Gamma^2E=ker(E\otimes E\to\Lambda E)$ are the multiples of the usual one. This argument works over any ground ring, and shows that the two bundles are isomorphic only if $2$ is invertible.