I shall show that the answer is no when $p=2$ (and it seems to me that a
somewhat more involved calculation will work for any $p$). We shall show that
there exists a vector bundle $\mathcal E$ such that $S^2\mathcal E$ is not
isomorphic to $\Gamma^2\mathcal E$ ($=(S^2\mathcal E^\ast)^\ast$).

Consider a vector bundle $\mathcal E$ which is an extension
$0\rightarrow\mathcal O_X\rightarrow\mathcal E\rightarrow\mathcal
O_X\rightarrow0$. We shall consider the structure group for such extensions.
Hence we assume $\mathcal E$ has a basis $e_1=1\in\mathcal O_X$ and $e_2$
(mapping to $1\in\mathcal O_X$). Adapted base changes have the form $e_1\mapsto
e_1$ and $e_2\mapsto e_2+he_1$. On the basis $e_1^2,e_1e_2,e_2^2$ of
$S^2\mathcal E$ we get $e_1^2\mapsto e_1^2$, $e_1e_2\mapsto e_1e_2+he_1^2$ and
$e_2^2\mapsto e_2^2+h^2e_1^2$. This gives us (globally) an exact sequence
$$
0\rightarrow S^2\mathcal O_Xe_1^2\rightarrow S^2\mathcal E\rightarrow\mathcal
O_Xe_1e_2\bigoplus\mathcal O_Xe_2^2\rightarrow0
$$
and if $g$ is the extension class for $\mathcal E$ we get that the extension
class for this extension is $ge_1e_2+F(g)e_2^2$ (with $F$ the Frobenius). Hence
if $g,F(g)\in H^1(X,\mathcal O_X)$ are linearly independent over the base field $k$,
then $S^2\mathcal O_X\rightarrow S^2\mathcal E$ induces an isomorphism on global
sections so that $H^0(X,S^2\mathcal E)=k$.

On the other hand, we have a basis $e_1^{(2)},e_1e_2,e_2^{(2)}$ for
$\Gamma^2\mathcal E$ ($(-)^{(2)}$ denoting the divided power). This gives
$e_1^{(2)}\mapsto e_1^{(2)}$, $e_1e_2\mapsto e_1e_2$ and $e_2^{(2)}\mapsto
e_2^{(2)}+he_1e_2+h^2e_1^{(2)}$ so that we get an exact sequence
$$
0\rightarrow \mathcal O_Xe_1^{(2)}\bigoplus\mathcal O_Xe_1e_2\rightarrow
\Gamma^2\mathcal E\rightarrow\mathcal O_Xe_2^{(2)}\rightarrow0.
$$
(This can also be seen by dualising the argument for $S^2$.) Hence
$H^0(X,\Gamma^2\mathcal E)$ is at least $2$-dimensional.

It is easy enough to get examples where $g\in H^1(X,\mathcal O_X)$ is linearly
independent from $F(g)$. One may for instance take an ordinary genus $2$ curve
for which there is a basis $a,b\in H^1(X,\mathcal O_X)$ with $F(a)=a$ and
$F(b)=b$ and let $g=a+\lambda b$ with $\lambda\notin\mathbb F_p$ or
supersingular but not superspecial genus $2$ curve which has $a\in
H^1(X,\mathcal O_X)$ with $a,Fa$ a basis and $F^2a=0$.

**Comment**: The example looks very special but in some sense it is exactly
the beviour under extensions that is the problem. Indeed, given a vector bundle
$\mathcal E$ we may consider the complete flag space of $\mathcal E$ over
$X$. If $S^m\mathcal E$ and $\Gamma^m \mathcal E$ were isomorphic over the flag
space, then we can push down such an isomorphism and get an isomorphism over
$X$. Hence, we may assume that $\mathcal E$ has a complete flag. As was noticed
in the question, if $\mathcal E$ is actually a direct sum of line bundles we
have an isomorphism.

**Addendum**: The example above is on a curve and the problem is invariant
under twisting (by a line bundle). Every rank $2$ vector bundle on a curve can up
to twisting be generated by three sections. Hence the vector bundle (again up to
twisting) is the pullback by a map from the curve to $\mathbb P^2$ of the
tautological rank $2$-vector quotient bundle (dual of the tautological
sub-bundle). Hence for that bundle on $\mathbb P^2$ we have that the symmetric
and divided squares are non-isomorphic. This bundle is, up to a twist, the
tangent bundle of $\mathbb P^2$ and hence the same is true for the tangent
bundle. As graded modules over the polynomial ring $k[x,y,z]$ we have explicit
presentations of these two modules. In principle we should be able to directly
show that they are non-isomorphic and in practice I guess Macaulay should be
able to do it.

**Addendum 1**: In fact it can. (**Previous code had incorrect presentation for divided sqaure.**.) Here is the code:

R = ZZ/2[x,y,z];
F = R^6;
Fzz = F_0;
Fxx = F_1;
Fxy = F_2;
Fxz = F_3;
Fyy = F_4;
Fyz = F_5;
Isym = R*(x*Fxx+y*Fxy+z*Fxz)+R*(x*Fxy+y*Fyy+z*Fyz)+R*(x*Fxz+y*Fyz+z*Fzz);
Idiv = R*(y*Fxy+z*Fxz)+R*(x*Fxy+z*Fyz)+R*(x*Fxz+y*Fyz)+R*(x^2*Fxx+y^2*Fyy+z^2*Fzz+x*y*Fxy+x*z*Fxz+y*z*Fyz);
Sym = sheaf (F/Isym);
Div = sheaf (F/Idiv);
print ({Sym,Div} / (s -> hilbertPolynomial(HH^0(s(*)))));
print ({Sym,Div} / (s -> poincare(HH^0(s(*)))));
print rank Hom(Sym,Div);
{3*P + 3*P , 3*P + 3*P }
1 2 1 2
-1 2
{6 - 3T, T + 3 - T }
1

Here I first verify that the Hilbert polynomials are the same (as they should having the same Chern classes) and then show that their Hilbert functions are different. Finally, I show that there is only one non-zero homomorphism $\mathrm{Sym}\to\mathrm{Div}$. As we already know one such, namely the one that is given by multiplication in the divided power algebra, and it is *not* an isomorphism we conclude that they are not isomorphic.
(Note that I put the partial derivatives $\partial/\partial x,\dots$ in degree $0$ instead of degree $-1$ where they belong.)