A "$n$-order matrix" $T\in M_n(\mathbb F_2)$ is a matrix such that there exists a partial ordered relation $\leq_T\subset [1,n]^2$ such that :
$T_{ij}=1\Leftrightarrow i\leq_T j$
(where $T_{ij}$ is the $i,j$ coefficient of $T$)
A "$n$-graph matrix" $S\in M_n(\mathbb F_2)$ is a symmetric matrix such that $S_{ii}=0$ for all $i\in [1,n]$
(where $G^t$ is the transposed matrix of $G$)
Is it true that for any "$n$-graph matrix" $S$ there exists a "$n$-order matrix" $T$ s.t. $S=T+T^t$ ?
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.
Here is a counting argument showing that almost all $n$-graph matrices are counterexamples.
There are $2^{(1/2+o(1))\cdot n^2}$ graphs on $n$ vertices, but only $2^{(1/4+o(1))\cdot n^2}$ partial orders on $n$ elements; see these papers by Kleitman and Rothschild:
Asymptotic enumeration of partial orders on a finite set
The number of finite topologies
The main reason for the separation is that the Hasse diagram is triangle-free, and there are only $2^{(1/4+o(1))\cdot n^2}$ triangle-free graphs on $n$ vertices (see e.g. https://arxiv.org/abs/1409.8123 for references).
As pointed out by others, the answer is no. In fact, the graphs that can be obtained this way are exactly the Comparability Graphs. This graph class is very well studied and well understood. It is also a subclass of perfect graphs, which are known to not allowing any induced off cycle with at least 5 vertices (such as the one in Nik Weaver's answer). Hence, any graph that consists of an odd cycle (or contains one as induced subgraph) cannot be obtained this way.
