Are supervector spaces the representations of a Hopf algebra?

The answer is yes, but comultiplication is not what you change. The symmetrizer (or braiding as you call it) is given by an $R$-element in $H \otimes H$ that makes $H$ into a triangular Hopf algebra. (The Wikipedia article says quasitriangular; triangular means that plus that the modified switching map $\widetilde{R}$ is an involution.)

The $R$ element is the correction to the usual switching map $x \otimes y \mapsto y \otimes x$. You can solve for it directly in terms of the usual description of the switching map on supervector spaces, and worry later about what axioms it satisfies. Let $1$ and $a$ be the group elements of $H$ and let $\epsilon$ and $\sigma$ be the dual vectors on $H$ which are the trivial and sign representations. Then we want $$(\epsilon \otimes \epsilon)(R) = 1 \quad (\epsilon \otimes \sigma)(R) = 1 \quad (\sigma \otimes \epsilon)(R) = 1 \quad (\sigma \otimes \sigma)(R) = -1,$$ because that is the correction in the modified switching map $x \otimes y \mapsto (-1)^{|x||y|} y \otimes x$. You can solve the linear system of equation to obtain $$R = \frac{1 \otimes 1 + a \otimes 1 + 1 \otimes a - a \otimes a}{2}.$$

As a monoidal category, supervector spaces are the same as $\mathbb{Z}/2\mathbb{Z}$ representations, so you can think of them as representations of the Hopf algebra $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$.

However, the symmetric structure is different, so the thing you need to do is make $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$ into a quasi-triangular Hopf algebra in an interesting way, that is, change things so that the map from $V\otimes W\to W\otimes V$ isn't just flipping, it's flipping followed by an element of $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]\otimes \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$ which is called the R-matrix. I'll leave actually writing this element as an exercise to the reader, but it's uniquely determined by acting by -1 on sign tensor sign and by 1 on everything else.

You may want to have a look at the relevant Wikipedia article.

The correct Hopf algebra is not the group algebra $kC_2$ but its dual $kC_2^*$. If the characteristic of $k$ is not $2$ then accidentally $kC_2\cong kC_2^*$. Otherwise, Greg answered the question as your new $R$-element is no longer trivial $1\otimes 1$ but $1 \otimes - 2\delta_x \otimes \delta_x$ where $\delta_x$ is the delta function on the generator of $C_2$.