For the $i$th factor ${\mathbb H}^2$ in the product of hyperbolic planes, pick a complete geodesic $c_i$, $i=1,...,k$. The product 
$$
F=c_1\times ... \times c_k\subset X=\prod_{i=1}^k {\mathbb H}^2 
$$
is a $k$-flat, i.e. a totally-geodesic (although we do not need this) isometrically embedded Euclidean subspace of 
dimension $k$. Let $(M,g)$ be a $d$-dimensional Riemannian manifold (no restrictions whatsoever). By the Nash isometric embedding theorem, there exists  an isometric embedding 
$$
f: (M,g)\to F,
$$
provided that $k$ is much larger than $d$. (Specifically, you can take any $k\ge (3d+11)/2$.) Then the composition of $f$ with the identity embedding $F\to X$, gives an isometric embedding $(M,g)\to X$. The image of this embedding (with the Riemannian metric induced from $X$) is a Riemannian submanifold of $X$ isometric to $(M,g)$. Now, if you wish, take $(M,g)$ to be say, the $d$-dimensional hyperbolic space (complete, simply-connected, constant curvature $-1$). 

Edit. It is possible that you are using a nonstandard notion of a Riemannian submanifold and what you really mean is a **totally geodesic** submanifold. Then the only complete negatively curved totally geodesic submanifolds of $X$ (of dimension $\ge 2$) are hyperbolic planes.