(6 October 2023) I'll leave the original argument below because it seems that many people liked it, but, in fact, it wanders around and introduces a lot of unnecessary information, which obscures the essential point, so here is a cleaned up version:
The answer 'yes' when $n\le 3$ but 'no' when $n\ge4$: Let $V = \mathbb{R}^n$, so $W = S^2(V^*)$ and the quadratic forms on $W$ are $S^2(W^*) = S^2\bigl(S^2(V^*)^*\bigr) = S^2(S^2(V))$ (since $S^2(V^*)$ is canonically isomorphic to $S^2(V)^*$). The (linear) 'multiplication map' $\mu:S^2(S^2(V))\to S^4(V)$ is surjective. Thus, each quadratic form on $W$ defines a quartic function on $V^*$, and every quartic function on $V^*$ is of the form $\mu(Q)$ for some $Q\in S^2(W^*)$.
The question is whether every non-negative quartic function on $V^*$ is of the form $\mu(Q)$ for some non-negative quadratic form $Q \in S^2(W^*)$. This is equivalent to asking whether every non-negative quartic function on $V^*$ is a sum of squares of quadratic functions on $V^*$.
However, it is known that for $n\ge4$, there exist non-negative quartic polynomials $P$ on $V^*$ that cannot be written as a sum of squares of quadratic polynomials on $V^*$, while, for $n\le 3$, any non-negative quartic polynomial $P$ on $V^*$ can be written as a sum of squares of quadratic polynomials.
Original Argument:
The answer 'yes' when $n\le 3$ but 'no' when $n\ge4$. Here is why:
Let $V = \mathbb{R}^n$, so $W = S^2(V^*)$ and the quadratic forms on $W$ are $S^2(W^*) = S^2\bigl(S^2(V^*)^*\bigr)$. It is known that there is a canonical $\mathrm{GL}(V)$-invariant exact sequence
$$
0\longrightarrow \Lambda^4(V)\longrightarrow S^2\bigl(\Lambda^2(V)\bigr)\longrightarrow S^2\bigl(S^2(V^*)^*\bigr)\longrightarrow S^4(V)\longrightarrow 0,
$$
thus, there is a $\mathrm{GL}(V)$-module $K(V)$ such that
$$
S^2\bigl(\Lambda^2(V)\bigr) = \Lambda^4(V)\oplus K(V)
\qquad\text{and}\qquad
S^2\bigl(S^2(V^*)^*\bigr) = S^4(V)\oplus K(V).
$$
The module $K(V)$ is $\mathrm{GL}(V)$-irreducible and of dimension $n^2(n^2{-}1)/12$. It is exactly the set of quadratic forms $Q$ on $W$ that vanish on all of the symmetric bilinear forms on $V$ of the form $B(v,w) = \ell(v)\ell(w)$ for some $\ell\in V^*$. (The quadratic forms with this latter property must be some $\mathrm{GL}(V)$-invariant submodule of $S^2\bigl(S^2(V^*)^*\bigr)$, and it clearly does not contain $S^4(V)$.)
Suppose that $Q$ be nonnegative on all of the rank 1 elements of $W$. This is equivalent to the condition that, when we write $Q = Q' + Q''$, with $Q'\in S^4(V)$ and $Q''\in K(V)$, then $Q'$, when considered as a quartic polynomial on $V^*$, be a nonnegative quartic polynomial.
If $\tilde Q$ were a non-negative quadratic form on $W$, then it would be a sum of squares of linear functions on $W$, and hence, if $\tilde Q - Q$ lay in $K(V)$, it would follow that $Q'$ would be a sum of squares of quadratic functions on $V^*$.
However, it is known that for $n\ge4$, there exist non-negative quartic polynomials $Q'$ on $V^*$ that cannot be written as a sum of squares of quadratic polynomials on $V^*$.
Thus, for $n\ge 4$, the answer to the OP's question is 'no'.
Meanwhile, when $n\le 3$, every non-negative quartic polynomial in $n$ variables is a sum of squares of quadratic polynomials in those $n$ variables, so the above analysis shows that, when $Q = Q' + Q''$ where $Q'\in S^4(V)$ is a nonnegative function on $V^*$ and $Q''\in K(V)$, there does exist a nonnegative quadratic $\tilde Q\in S^2\bigl(S^2(V^*)^*\bigr)$ such that $\tilde Q - Q$ lies in $K(V)$.
