Suppose we already know that the process is continuous with probability one, so that the process takes values in the Banach space $X = C([0,1])$ with distribution $\mathbb P$. The covariance operator $C : X^* \to X$ is then a map from the dual space $X^*$ to $X$. The support of the Gaussian measure $\mathbb P$ is then the closure of the image of $X^*$ under $C$: $$\operatorname{supp} \mathbb P = \overline{ CX^* }.$$ (This is the main theorem of [Vakhania 1975])

Now let's construct the Cameron-Martin space. The operator $C$ defines an inner product on the dual space $X^*$ by $$\langle f, g \rangle = f(Cg)$$ for $f, g \in X^*$. The space $X^*$ isn't necessarily closed under the topology induced by the inner product, so let $H$ be the Hilbert space completion, and let $\iota : X^* \hookrightarrow H$ be the inclusion map. Define a map $\iota^* : H \hookrightarrow X$ first on the dense subspace $\iota X^* \subseteq H$ by $$\iota^*(\iota f) = Cf,$$ and extend continuously to all of $H$. Thus the covariance operator factors as $C = \iota^* \circ \iota$.

The Hilbert space $\iota^* H$ is a subspace of $X$, and is called the Cameron-Martin space of the process. Interpreting this in the context of the support of the Gaussian measure $\mathbb P$, we have $$\operatorname{supp} \mathbb P = \overline{\iota^* H},$$ so that the closure of the Cameron-Martin space (with respect to the original norm of $X$) is exactly the support of $\mathbb P$.

I go into these ideas in more detail in Section 2 of my preprint [LaGatta 2010].

Suppose your covariance operator is an integral operator with kernel $c(s,t)$, called the covariance function of the process. That is, if $\mu$ is a Radon measure on $[0,1]$, then $$(C\mu)(s) = \int_0^1 c(s,t) \, d\mu(t).$$

If we write $c_s(t) = c(s,t)$, then the support of $\mathbb P$ is the closure of the span of the functions $c_s$ in $C([0,1])$. So to answer your question: if you know that the closure $\overline{\operatorname{span}\{c_s\}}$ is a space with regularity property $P$, then the process $\xi_t$ satisfies property $P$ with probability one.