I get the definition of a Hamiltonian matrix from Wikipedia and this article and they both agree.
Is the Quantum Mechanical Hamiltonian operator when expressed as a matrix, a Hamiltonian matrix, or are they two separate things?
Because the matrix of the operator in this link does not seem to fit the definition of a Hamiltonian matrix, and nor did a few others.
They are two separate things. The matrix $$ \begin{bmatrix}0&I_{n}\\-I_{n}&0\end{bmatrix} $$ is the standard matrix representation of the symplectic form $\Omega$ on a symplectic vector space, while the Hamiltonian operator, as defined in quantum mechanics, is associated to the energy of a physical system: if a physical system is in some state, a measurement of energy will be one of the eigenvalues of the Hamiltonian operator.
The Hamiltonian matrix associated with a Hamiltonian operator $H$ is simply the matrix of the Hamiltonian operator in some basis, that is, if we are given a (countable) basis $\{|i\rangle\}$, then the elements of the Hamiltonian matrix are given by $$ H_{ij}=\langle i|H|j\rangle. $$ Notice that this has nothing to do with the definition of Hamiltonian matrix you have seen in the article in Wikipedia.
Although the concepts of Hamiltonian operator and symplectic form are different, they are closely connected. Let me illustrate this fact by considering classical Hamiltonian mechanics, that is, the dynamics of $n$ particles on the phase space with coordinates given by the positions and momenta $(q,p)=(q_1,\ldots,q_n,p_1,\ldots,p_n)$ and Hamiltonian function $H$. The Hamilton's equations of motion are given by $$ \frac{dq_i}{dt}=\frac{\partial H}{\partial p_i},\\ \frac{dp_i}{dt}=-\frac{\partial H}{\partial q_i}. $$ Defining $\eta=(q,p)$, the equations of motion can be written as $$ \frac{d\eta}{dt}=\Omega\nabla H. $$
If you understand some differential geometric language, I have given an answer that shows the relation between the symplectic form and Hamiltonian vector fields on symplectic manifolds.