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][1] 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 $K(V)$. 

  [1]: https://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem