The Gaussianity assumption mentioned in the answer of Joris is a particular case of a more general condition:

The family of random variables $\{\langle X, \varphi \rangle^2 \mid \varphi \in S\}$ is uniformly integrable, where $S$ is the set of linear functionals of the form $\langle X, \varphi \rangle := \sum_{k=1}^n \varphi_k X(t_k)$ whose $\mathcal{H}(c)^\ast$-norm is $\le 1$ (i.e. $\forall \xi \in \mathcal{H}(c): \sum_k \varphi_k \xi(t_k) \le \Vert \xi \Vert_{\mathcal{H}(c)}$).

In the Gaussian case this condition is satisfied because the $L^2(\Omega,\mathsf{P})$ norm of the linear functionals $\langle X, \varphi \rangle$ (which is the same as $\Vert \varphi \Vert_{\mathcal{H}(c)^\ast}$) bounds an $L^p(\Omega,\mathsf{P})$ norm for $p>2$. More generally, it is also satisfied for processes that depend polynomially on a Gaussian system, or on Bernoulli random signs, or basically on anything where hypercontractivity works.

In general the condition above is enough to conclude that the sample paths of $X$ do not lie in $\mathcal{H}(c)$ (unless it's finite-dimensional).

To prove this, note that by the Lebesgue's theorem, uniform integrability implies that the $L^2$ and $L^0$ topologies (the latter is convergence in probability) coincide on the space of linear functionals of $X$. Note that the $L^0$ topology won't change if we change the probability measure $\mathsf{P}$ by an equivalent measure $\mathsf{Q}$.

Now argue towards a contradiction. Let $X$ be a random element of $\mathcal{H}(c)$. Then $\Vert X \Vert_{\mathcal{H}(c)}^2$ is a well-defined, finite random variable, so there exists a measure $\mathsf{Q}$, equivalent to $\mathsf{P}$, such that $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert_{\mathcal{H}(c)}^2 < \infty$. On the other hand, $\mathsf{E}_{\mathsf{Q}} \Vert X \Vert^2_{\mathcal{H}(c)}$ is the trace of the covariance operator of $X$ in $\mathcal{H}(c)$ (computed for the measure $\mathsf{Q}$), so the covariance must be trace class, and therefore compact. Thus there is a sequence of linear functionals $\varphi_n \in S$, such that $\Vert \varphi_n \Vert_{\mathcal{H}(c)^\ast} = 1$, and $\langle \varphi_n, X \rangle \to 0$ in $L^2(\mathsf{Q})$. Thus we also have $\langle \varphi_n, X \rangle \to 0$ in $L^0(\mathsf{P})$, and, by uniform integrability, also in $L^2(\mathsf{P})$. This contradicts the fact that $\Vert \langle \varphi, X \rangle \Vert_{L^2(\mathsf{P})} = \Vert \varphi \Vert_{\mathcal{H}(c)^\ast} = 1$.