Here is a counterexample. Let $k$ be a field of characteristic $2$, let $R=k[x,y,z]$, and let $M$ be the ideal $(x,y,z)\subset R$. Note that $xy\otimes z=x\otimes yz=xz\otimes y=z\otimes xy$ in $M^{\otimes 2}$, so $xy\otimes z\in (M^{\otimes 2})^{\Sigma_2}$. However, I claim that $xy\otimes z$ is not in the span of the image of $\operatorname{Sym}:M^2\to M^{\otimes 2}$.
To prove this, it suffices to show $xy\otimes z$ is not in the span of the degree $\leq 3$ parts of tensors of the form $a\otimes a$ or $a\otimes b+b\otimes a$, for $a,b\in M$. Moreover, $a$ and $b$ can be assumed to be monomials (if they are sums of monomials, then $a\otimes a$ and $a\otimes b+b\otimes a$ expand out into a sum of such terms formed by monomials). Note also that if $a$ is a monomial of degree $2$ and $b$ is a monomial of degree $1$, then $a\otimes b=b\otimes a$ by a manipulation similar to the one used above to show $xy\otimes z=z\otimes xy$. Since $k$ has characteristic $2$, this means $a\otimes b+b\otimes a$ will be $0$ in that case. Similar reasoning also shows that if $a$ and $b$ have degree $1$, then any degree $3$ multiple of $a\otimes b+b\otimes a$ will be $0$. So, we only need consider the case $a\otimes a$ where $a$ has degree $1$; that is, we must show that $xy\otimes z$ is not in the submodule generated by $x\otimes x,y\otimes y,$ and $z\otimes z$. This is easy: there is a homomorphism $M^{\otimes 2}\to R^{\otimes 2}\cong R$ which sends $xy\otimes z$ to $xyz$, which is not in the submodule generated by $x^2,y^2,$ and $z^2$.
On the other hand, it is true if $M$ is projective. First, it is easy to see it is true if $M$ is free by picking a basis and considering the induced basis on $M^{\otimes n}$, which is permuted by the action of $\Sigma_n$. Now suppose $M$ is projective, so it is a direct summand of a free module; say $F$ is free and $F=M\oplus N$. Note that the action of $\Sigma_n$ respects the direct sum decomposition $F^{\otimes n}=M^{\otimes n}\oplus K$ where $K$ is all the other terms you get by distributing the tensor product over the direct sum in $F=M\oplus N$. Also, if $v=(x,y)\in F^n$ with $x\in M^n$ and $y\in N^n$, then the $M^{\otimes n}$ component of $\operatorname{Sym}(v)$ is an integer multiple of $\operatorname{Sym}(x)$ (the multiple being because $\operatorname{Stab}(x)$ might be larger than $\operatorname{Stab}(v)$). It follows that the intersection of the span of the image of $\operatorname{Sym}:F^n\to F^{\otimes n}$ with $M^{\otimes n}$ coincides with the span of the image of $\operatorname{Sym}:M^n\to M^{\otimes n}$. But the span of the image of the span of the image of $\operatorname{Sym}:F^n\to F^{\otimes n}$ is all of $(F^{\otimes n})^{\Sigma_n}$ since $F$ is free, and so in particular it contains all of $(M^{\otimes n})^{\Sigma_n}$.
It is also true if $n!$ is a unit in $R$. In that case, given $x\in (M^{\otimes n})^{\Sigma_n}$, you can write $x$ as a sum of simple tensors $x=\sum_i t_i$, and then $$x=\frac{1}{n!}\sum_{\sigma\in\Sigma_n}\sigma(x)=\frac{1}{n!}\sum_i\sum_{\sigma\in\Sigma_n} \sigma(t_i)$$ and $\sum_{\sigma\in\Sigma_n} \sigma(t_i)$ is an integer multiple of an element of the image of $\operatorname{Sym}$ for each $i$.
