$\newcommand{\Sp}{\mathrm{Sp}\,}\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathrm{m}}$There seem to exist such examples and this does not contradict de Jong's argument because his proof only shows that the homomorphism $G:=\pi_1(\Sp C\langle x^{\pm 1}\rangle)\to H:=\pi_1(\Sp C\langle x\rangle)$ satisfies the following property: if an action of $H$ on a finite set becomes trivial when restricted to $G$ then it is trivial. Equivalently, the normal subgroup generated by the image of $G$ is equal to the whole of $H$.

Take $$A= C\langle x,y\rangle/(y^{p+1}-xy-p^{1/2})$$ The discriminant of this polynomial in $y$ is $-(p+1)^{p+1}p^{p/2}-p^px^{p+1}$ up to a unit and is invertible in $C\langle x\rangle$. Therefore, $C\langle x\rangle \to A$ is a finite etale extension.

To check that $\Sp A$ is connected it is enough to show that the polynomial $y^{p+1}-xy+p^{1/2}$ is irreducible in $C\langle x\rangle[y]$. If $y^{p+1}-xy-p^{1/2}=f_1(y)\cdot f_2(y)$ is a factorization into a product of monic polynomials then both $f_1,f_2$ have to lie in $\cO_C\langle x\rangle[y]$ so their reductions modulo the maximal ideal $\fm_C\subset\cO_C$ provide a factorization of $(y^{p}-x)y$. Hence, we may and will assume that $\deg f_1=1$ and $y^{p+1}-xy-p^{1/2}$ has a root $f(x)\in \cO_C\langle x\rangle$. A root has to have $f(0)^{p+1}=p^{1/2}$ and, taking the derivative of the equation $f(x)^{p+1}-xf(x)+p^{1/2}=0$ we also get $(p+1)f'(x)f(x)^p-f(x)-xf'(x)=0$ so $f'(0)=\frac{1}{p+1}f(0)^{1-p}\not\in \cO_C$ which is a contradiction. Therefore, $\Sp A\to D$ is a connected finite etale cover.

On the other hand, this cover splits over $C\langle x,x^{-1}\rangle $: we can find a root of $y^{p+1}-xy-p^{1/2}=0$ by the Hensel's lemma arguing by induction on $n$: suppose that $y_n\in \cO_C/p^{n/2}[x^{\pm 1}]$ is a root of this equation modulo $p^{n/2}$ that reduces to $0$ mod $p^{1/2}$. To lift it over $p^{(n+1)/2}$ it is enough to be able to supply, for any given $a\in\cO_C/p^{1/2}[x^{\pm 1}]$, an element $z$ such that $(p+1)y_n^p\cdot z-x\cdot z=a$ in $\cO_C/{p^{1/2}}[x^{\pm 1}]$. This is possible because $y_n^p-x$ is a sum of a nilpotent and invertible element, hence is invertible.

This example is motivated by the behavior of the $p$-torsion in a family of elliptic curves: over the special fiber it is a union of irreducible components such that over the ordinary locus the canonical subgroup $\ker F$ provides a separate connected component that collides with the other component over the special fiber. By the theory of the canonical subgroup the splitting over the ordinary locus can be lifted to the generic fiber. In the example above $x=0$ plays the role of "supersingular" locus.

$\newcommand{\Sp}{\mathrm{Sp}\,}\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathrm{m}}$There seem to exist such examples and this does not contradict de Jong's argument because his proof only shows that the homomorphism $G:=\pi_1(\Sp C\langle x^{\pm 1}\rangle)\to H:=\pi_1(\Sp C\langle x\rangle)$ satisfies the following property: if an action of $H$ on a finite set becomes trivial when restricted to $G$ then it is trivial. Equivalently, the normal subgroup generated by the image of $G$ is equal to the whole of $H$.

Take $$A= C\langle x,y\rangle/(y^{p+1}-xy-p^{1/2})$$ The discriminant of this polynomial in $y$ is $-(p+1)^{p+1}p^{p/2}-p^px^{p+1}$ up to a unit and is invertible in $C\langle x\rangle$. Therefore, $C\langle x\rangle \to A$ is a finite etale extension.

To check that $\Sp A$ is connected it is enough to show that the polynomial $y^{p+1}-xy+p^{1/2}$ is irreducible in $C\langle x\rangle[y]$. If $y^{p+1}-xy-p^{1/2}=f_1(y)\cdot f_2(y)$ is a factorization into a product of monic polynomials then both $f_1,f_2$ have to lie in $\cO_C\langle x\rangle[y]$ so their reductions modulo the maximal ideal $\fm_C\subset\cO_C$ provide a factorization of $(y^{p}-x)y$. Hence, we may and will assume that $\deg f_1=1$ and $y^{p+1}-xy-p^{1/2}$ has a root $f(x)\in \cO_C\langle x\rangle$. A root has to have $f(0)^{p+1}=p^{1/2}$ and, taking the derivative of the equation $f(x)^{p+1}-xf(x)+p^{1/2}=0$ we also get $(p+1)f'(x)f(x)^p-f(x)-xf'(x)=0$ so $f'(0)=\frac{1}{p+1}f(0)^{1-p}\not\in \cO_C$ which is a contradiction. Therefore, $\Sp A\to D$ is a connected finite etale cover.

On the other hand, this cover splits over $C\langle x,x^{-1}\rangle $: we can find a root of $y^{p+1}-xy-p^{1/2}=0$ by the Hensel's lemma arguing by induction on $n$: suppose that $y_n\in \cO_C/p^{n/2}[x^{\pm 1}]$ is a root of this equation modulo $p^{n/2}$ that reduces to $0$ mod $p^{1/2}$. To lift it over $p^{(n+1)/2}$ it is enough to be able to supply, for any given $a\in\cO_C/p^{1/2}[x^{\pm 1}]$, an element $z$ such that $(p+1)y_n^p\cdot z-x\cdot z=a$ in $\cO_C/{p^{1/2}}[x^{\pm 1}]$. This is possible because $y_n^p-x$ is a sum of a nilpotent and invertible element, hence is invertible.

This example is motivated by the behavior of the $p$-torsion in a family of elliptic curves: given such family $\mathcal{E}\to S$ over a $p$-adic formal scheme $S$ over $\cO_K$ let $S_0(p)\to S_K$ be the etale covering of the generic fiber parametrizing $1$-dimensional $\mathbb{F}_p$ subspaces in $\mathcal{E}_K[p]$. When restricted to the generic fiber of the ordinary locus $S_{ord}$ this covering admits a section by the theory of canonical subgroup but it need not have one over the whole of $S_K$. In other words the restriction of the representation $\pi_1(S_K)\to GL_2(\mathbb{F}_p)$ to $\pi_1(S_{ord})$ lands inside a Borel subgroup.

Using Katz-Mazur-Drinfeld integral model the etale cover $S_0(p)\to S_K$ extends to a flat cover of $S$ itself and the special fiber splits into two irreducible components exactly as in the example above.