I've stumbled across the above question. Usually, I don't spend my time on such problems, though it's a nice question in my opinion.
My idea:
Case $p=2$ is easy, follows from $x^2 \equiv 2 \;\;\; (\text{mod} \;\; 4)$ has no integer solution.
Assume the contrary and $p$ is an odd prime number. Since, $A+A^t$ is symmetric, then is has only real eigenvalues, so if $\det(xI-(A+A^t)) \in \mathbb{Z}[x]$ is the characteristic polynomial, it factors $(x-\lambda_1) \cdots (x-\lambda_p)$ where $\lambda_i$'s are real numbers. By rational root test, if any of $\lambda_i$ is rational, then it will be an integer. Therefore, if we show that non of them is irrational (hence, all $\lambda_i$ are integers), we may then conclude as follows;
$\sum_i \lambda_i=-tr(A+A^t)=2k$ where $k \in \mathbb{Z},$ and $\lambda_1 \cdots \lambda_p=-p.$ Now, $p$ is odd, then wlog,
$\lambda_p=-p$ and $\lambda_1=\cdots=\lambda_{2t}=-1$ and $\lambda_{2t}=\cdots=\lambda_{p-1}=1,$ but $\sum_i \lambda_i=-2t+p-1-2t-p$ which is odd,
$\lambda_p=p$ and $\lambda_1= \cdots=\lambda_{2t+1}=-1$ and $\lambda_{2t+1}=\cdots=\lambda_{p-1}=1,$ but again $\sum_i \lambda_i=-2t-1+p-1-2t-1+p$ which is odd,
hence contradiction.
Now, we're left to show that (in my words)
Lemma: Given $p$ integers and irrational numbers, where $p$ is an odd prime number. We have that
$\sum_i\lambda_i=2k \in \mathbb{Z}, $
$\sum_{i
$.$
$.$
$.$
$\lambda_1\cdots \lambda_p=-p \in \mathbb{Z}.$
Then, non of the $\lambda_i$ is irrational.
An idea to prove the lamma: Since any symmetric polynomial in $p$ variables $\lambda_i$ with integer coefficient can be expressed in terms of elementary symmetric polynomials above, then any symmetric (polynomial with integer coefficient) expression of $\lambda_i$'s will be an integer.
Is it possible to conclude from here that non of the $\lambda_i$ is irrational?
P.S. I'd appreciate any other ideas for solving it.