Even with $0$'s on the diagonal it's still false. Consider the matrix $S = \left[\begin{matrix}0&1&0&0&1\cr 1&0&1&0&0\cr 0&1&0&1&0\cr 0&0&1&0&1\cr 1&0&0&1&0\end{matrix}\right]$. Suppose $S = T + T^t$ for some order matrix $T$. Then this order must have either $1 < 2$ or $2 < 1$ since $s_{12} = 1$. Wlog say $1 < 2$. Similarly we must have $2 < 3$ or $3 < 2$.

Suppose $2 < 3$. Then $1 <2 < 3$, and hence $t_{13} = 1$, but $s_{13} = 0$ so this implies that $t_{31} = 1$, i.e., $3 < 1$. If a "partial order" is antisymmetric this is impossible, but even if you allow $1 < 3 < 1$ we then get $2 < 3 < 1$, so that $t_{12} = t_{21} = 1$ and this contradicts $s_{12} = 1$. So $2 < 3$ is impossible.

We have shown that $1 < 2$ implies $3 < 2$, and similarly $2 > 3$ implies $4 > 3$, and then $3 < 4$ implies $5 < 4$.

But what is the relation between $1$ and $5$? Since $s_{15} = 1$ we need either $1 < 5$ or $5 < 1$. Wlog say $1 < 5$. Then $1 < 5 < 4$ so $t_{14} = 1$, but since $s_{14}= 0$ this forces $t_{41} = 1$, i.e., $4 < 1$. Again this contradicts antisymmetry, but even if you drop antisymmetry it forces $5 < 4 < 1$ so that $t_{51} = t_{15} = 1$ and this contradicts $s_{15} = 1$. So $S = T + T^t$ is impossible.