The answer is 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 $$ S^2\bigl(\Lambda^2(V)\bigr) = \Lambda^4(V)\oplus K(V) \qquad\text{and}\qquad S^2\bigl(S^2(V^*)^*\bigr) = S^4(W)\oplus K(V). $$ 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 $S^4(W)$.)
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'. Whether the answer is 'no' for $n=3$ is a good question.