Hardness of a constrained quadratic maximization
(I have found the answer to my question, so I am sharing a sketch here).

The quadratic maximization can be shown to be NP-hard by a reduction from the following problem:


Given $k$-partite graph $G=(V_{1}, \dots, V_{k}, E)$, does $G$ contain a $k$-clique. 


Note that the latter problem is a (seemingly) special case of the max-clique problem restricted to a particular family of graphs. It can however be shown that it is also NP-complete by a reduction from the general max-clique problem itself (See here).

Let $\mathbf{A}$ denote the adjacency matrix of $G$.
Let $S$ be an arbitrary set of $k$ vertices and $\mathbf{A}_{S}$ denote the principal submatrix corresponding to $S$, i.e. $\mathbf{A}_{S}$ is the $k\times k$ adjacency of the graph induced by $S$.
If the vertices in $S$ form a $k$-clique in $G$, then all off-diagonal entries of $\mathbf{A}_{S}$ are equal to $1$, and its principal eigenvalue $\lambda_{1}(\mathbf{A}_{S}) = k-1$.
In any other case, that is, if $S$ is not a $k$-clique, then $\lambda_{1}(\mathbf{A}_{S}) &lt; k-1$.
Finally, note that since $G$ is $k$-partite, a $k$-clique (if one exists) will contain a single vertex from each of the sets $V_{i}$, $i=1,\dots,k$.

Let $\mathbf{A}^{\prime} = \mathbf{A} + |\lambda_{n}(\mathbf{A})|\cdot\mathbf{I}$, where $\lambda_{n}(\mathbf{A})$ is the smallest eigenvalue of $\mathbf{A}$; 
$\mathbf{A}^{\prime}$ is a PSD matrix.
Further, consider $k$ disjoint groups of variables corresponding to the sets $V_{1}, \dots, V_{k}$ of $G$.
We solve the quadratic maximization with input $\mathbf{A}^{\prime}$ and the specified groups of variables.
The maximum objective value will be equal to $k-1 + |\lambda_{n}(\mathbf{A})|$ if and only if $G$ contains a $k$-clique.