Since the Cantor set $K$ has Hausdorff dimension $\log2/\log 3<1$, it is a removable set for bounded analytic functions, and so, as you say, there is no bounded analytic function outside of $K$. But it does not mean that there are no bounded harmonic functions outside of $K$. Actually, any point of $K$ is regular for the Dirichlet problem. So, choosing any non-constant continuous function $u$ on $K$, there exists a unique harmonic function $f$ in $\overline{\mathbb C}\setminus K$ such that $$ \forall\zeta\in K,~\lim_{z\to\zeta}f(z)=u(\zeta). $$ Then $f$ is a non-constant bounded harmonic function in $\overline{\mathbb C}\setminus K$. Note that $K$ is of positive capacity so there is no reason that $f$ can be extended to $\mathbb C$ (and actually it cannot be extended).

Since the Cantor set $K$ has Hausdorff dimension $\log2/\log 3<1$, it is a removable set for bounded analytic functions, and so, as you say, there is no bounded analytic function outside of $K$. But it does not mean that there are no bounded harmonic functions outside of $K$. Actually, any point of $K$ is regular for the Dirichlet problem. So, choosing any non-constant continuous function $u$ on $K$, there exists a unique harmonic function $f$ in $\overline{\mathbb C}\setminus K$ such that $$ \forall\zeta\in K,~\lim_{z\to\zeta}f(z)=u(\zeta). $$ Then $f$ is a non-constant bounded harmonic function in $\overline{\mathbb C}\setminus K$. Note that $K$ is of positive capacity so there is no reason that $f$ can be extended to $\mathbb C$ (and actually it cannot be extended).

A proof that any point of $K$ is regular for the Dirichlet problem, that uses Wiener criterion, is given p.100 of

J.B. Garnett, D.E. Marshall, Harmonic measure. New Mathematical Monographs 2. Cambridge University Press, Cambridge, 2005.