Can a connected planar compactum minus a point be totally disconnected?

Being planar has nothing to do with the problem. Suppose a totally disconnects $X$ and choose $b$ different from $a$. By passing to a sub continuum, assume that no proper sub continuum contains both $a$ and $b$. Take non empty disjoint open sets $U$ and $V$ whose union is $ X\sim a$. WLOG $b$ is in $U$, and observe that $U\cup \{a\}$ is closed and connected.

Let denote by $U_n\subset \mathbb R^2$ a sequence of open bounded neigborhoods of $X$, so that $$U_{n+1}\subset U_n\ \ \text{and}\ \ \bigcap_n U_n=X.$$ We can assume that all $U_n$ are connceted and therefore path-connected. Coose a point $p\in X$ distict from $x$ and consider a sequence of paths $\gamma_n$ in $U_n$ from $p$ to $x$. Fix $\epsilon>0$ such that $\epsilon<|p-x|$. For each path choose the smalest value $t_n\in[0,1]$ so that $|\gamma_n(t_n)-x|=\epsilon$. The image $Z_n=\gamma([0,t_n])$ is connected compact set. Let $Z$ be a Hausdorff limit of a subsequence of $Z_n$. Note that $Z$ is a compact connected subset of $X$. Clearly, $Z\not\ni x$ and it contains at least two points; a contradiction

Two great answers have already been given, and I don't claim to add much, but here is something anyway.

A totally disconnected locally compact Hausdorff space has a basis of clopen sets, according to Proposition 3.1.7 of Arhangel'skii and Tkachenko, for example. A closed set in $X-\{a\}$ need not be closed in $X$, but if $X$ is a metric space then the clopen subsets of $X-\{a\}$ at positive distance to $a$ will be clopen in $X$. Thus if $X$ is a compact metric space with more than one point and $X-\{a\}$ is totally disconnected, then $X$ is not connected.