Yes, there is a combinatorial way to see this.
Firstly, it is much easier to show a baby version of this type of phenomenon. A while ago I read a nice blog post by Qiaochu explaining the identity
$$\int_{0}^1 (2\cos \pi x)^n (2\sin^2\pi x)dx=\begin{cases} 0 &\text{ if n is odd} \\
C_{n/2} & \text{ if n is even} \end{cases}$$
by interpreting this integral as $\int_{SU(2)}\chi_V(g)^nd\mu$, with respect to the normalized Haar measure, where $V$ is the defining representation. This in turn is the multiplicity of the trivial representation in $V^{\otimes n}$. Next, you make a graph with vertices corresponding to the irreducible representations of $SU(2)$, and add as many edges from $A$ to $B$ as the multiplicity of $A$ in $B\otimes V$. The irreducible representations of $SU(2)$ are the symmetric powers $\operatorname{Sym}^n(V)$, thus they are in bijection with the natural numbers. As a corollary of the fact that
$$\operatorname{Sym}^n(V)\otimes V\cong \operatorname{Sym}^{n-1}(V)\otimes \operatorname{Sym}^{n+1}(V)$$
we obtain that the graph is the half infinite line
$$0\leftrightarrow 1\leftrightarrow 2\leftrightarrow \cdots$$
and the desired multiplicity is the number of walks from zero to itself of length $n$, thus giving the answer in terms of the Catalan numbers.
Now for your question, the representation of $USp(2n)$ under consideration is the exterior algebra of the standard representation, $W$. So $V=\bigoplus_{k=0}^{2n}\Lambda^{k}(W)$ and your integral measures the multiplicity of the trivial representation in the representation $V^{\otimes r}$. A general combinatorial set up that keeps track of these types of multiplicities is that of crystal bases. It is much more technical but it is what ultimately brings the result in terms of enumerations of certain types of tableaux or non-intersecting lattice paths. Fortunately, the details of this specific problem are all worked out in the paper
Se-jin Oh, Travis Scrimshaw "Identities from representation theory" Discrete Math., 342(9):2493–2541, 2019 (arxiv)
