Edited in response to OP's edits of the question
$C$ is $n\times n$ covariance matrix of
(column) $n$-vectors $\mathbf x$ and $\mathbf y$. The $i$-$j$-th
entry in $C$ is $c_{i,j} = \text{cov}(X_i, Y_j)$. Now,
$E[X_i]=\hat{x}_i$ and $E[Y_j]=\hat{y}_j$ so that
$c_{i,j} = E[X_iY_j]-\hat{x}_i\hat{y}_j$. Then,
$$\begin{align*}
E[\mathbf x^TA\mathbf y] &= E\left[\sum_{i=1}^n\sum_{j=1}^n a_{i,j}X_iY_j\right]\\
&= \sum_{i=1}^n\sum_{j=1}^n a_{i,j}E[X_iY_j]\\
&= \sum_{i=1}^n\sum_{j=1}^n a_{i,j}(c_{i,j}+\hat{x}_i\hat{y}_j)\\
&= \hat{\bf x}\,^TA\hat{\bf y} + \sum_{i=1}^n\sum_{j=1}^n a_{i,j}c_{i,j}.
\end{align*}$$
If $C = \mathbf 0$ is the all-zeroes matrix,
then we get
$$E[\mathbf x^TA\mathbf y]
= \hat{\mathbf x}\,^TA\hat{\mathbf y} = E[\mathbf{x}]^TAE[\bf y]$$
as in Robert Israel's answer. (However, Robert has not as yet
revised his answer
in response to your edits that make the $X_i$ and $Y_j$ non-zero-mean
random variables and so $E[\mathbf{x}]^TAE[\mathbf y]$ does not equal
$0$ when $C = \mathbf 0$ as he says). Note that normality of $\mathbf x$ and/or $\mathbf y$ has nothing to do
with the matter.