Notice that $\langle A, B \rangle = \sum_{i,j} [A]_{ij} [B]_{ij}$. Let $E_{ij} = e_i e_j^T$, ie, the zero matrix except for a one in the $(i,j)$ position.
Suppose $\langle A, S \rangle = 0$ for all symmetric matrices, then it is true for $S=E_{ij}+E_{ji}$. This gives $\langle A, E_{ij} \rangle + \langle A, E_{ji} \rangle = 0$, which gives $[A]_{ij} + [A]_{ji} = 0$, from which it follows that $A = -A^T$.
Now suppose that $A$ is skew symmetric, and $S$ is symmetric. Then we can write $S = U^T + \Lambda + U$, where $\Lambda$ is diagonal, and $U$ is strictly upper triangular. Then $\langle A, \Lambda \rangle = 0$ because $[A]_{ii} = 0$. Since $\operatorname{tr} M = \operatorname{tr} M^T$ we also have $\langle A, U^T \rangle = \langle A^T, U \rangle$, and skew symmetry gives $\langle A^T, U \rangle = - \langle A, U \rangle$. Hence $\langle A, S \rangle = \langle A, U^T \rangle + \langle A, \Lambda \rangle - \langle A, U^T \rangle = 0$.
Well, in a way, yes. The real Jordan normal form theorem yields that every real matrix has an invariant subspace $W$ (i.e. $AW\subseteq W$) such that $1\le\dim W\le 2$. Since $\dim W^\perp +\dim W=n$ and, if $A$ is orthogonal, the orthogonal complement of an $A$-invariant subspace is $A$-invariant, there is an orthonormal basis $B=(b^1,\cdots, b^n)$ such that $B^{-1}AB=B^TAB=\begin{pmatrix}U &0\\ 0&U'\end{pmatrix}$, for some $U\in O(2)$ and $U'\in O(n-2)$ - or, repsectively, $O(1)$ and $O(n-1)$. Now, the eigenvalues of $A$ are either eigenvalues of $U$ or eigenvalues of $U'$. The form and eigenvalues of a $2\times 2$ orthogonal matrix can be calculated explicitly, and the rest can be done by induction.
That being said, as far as I know the real Jordan normal form theorem is proved by complexifying the endomorphism of $A$, using the machinery of Jordan normal form in $\Bbb C^n$, and then bringing it back to $\Bbb R^n$. So one could argue that this is just hiding the issue under the carpet.
Best Answer
No, the statement is not true. As a counterexample, consider $$ A = Q = V = \pmatrix{1&0\\0&1}, \quad U = \pmatrix{0&1\\1&0}. $$