Edit There was a major hole in the first version, fixed with the aid of a hint from hot_queen. See Below.
It goes the other way; CH implies that there is no such measure.
This follows from Ulam's theorem, which says that if $\mu$ is a finite measure defined on the power set of $\omega_1$ and $\mu$ vanishes on singletons then $\mu=0$.
Been wondering how to prove that; made this into an answer when I saw how simple it is. Say $\mu$ is a finite measure defined on the power set of $\omega_1$ which vanishes on the singletons. Consider the product measure $\mu\times\mu$ on $\omega_1\times\omega_1$. Let $A=\{(\alpha,\beta)\in\omega_1\times\omega_1:\alpha<\beta\}$. Tonelli shows that $$0=\mu\times\mu(A)=\mu(\omega_1)^2.$$
In fact the same proof works assuming just that $\mu$ is $\sigma$-finite. On the other hand, if we define $\mu(E)$ for $E\subset\omega_1$ by $\mu(E)=0$ if $E$ is bounded and $\mu(E)=\infty$ otherwise then $\mu$ is a non-zero measure defined on the power set of $\omega_1$ which vanishes on singletons; hence you can't ignore the hypotheses in Tonelli's theorem.
Below: hot_queen pointed out that I neglected to show that $A$ was measurable (wrt to the product $\sigma$-algebra). This is not that hard once you see how to do it; I needed a hint.
There exist sets $I_{n,j}\subset\Bbb R$ such that $$\{(x,y)\in\Bbb R^2:x=y\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(I_{n,j}\times I_{n,j});$$for example $I_{n,j}=[j/n,(j+1)/n)$. Since $\omega_1$ is equivalent to a subset of $\Bbb R$ it follows that there exist $I_{n,j}\subset\omega_1$ such that$$\{(\alpha,\beta)\in\omega_1^2:\alpha=\beta\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(I_{n,j}\times I_{n,j}).$$Hence if $J_{n,j}$ is the set of countable ordinals greater than or equal to the smallest element of $I_{n,j}$ we have $$\{(\alpha,\beta)\in\omega_1^2:\alpha\ge\beta\}=\bigcap_{n\in\Bbb N}\bigcup_{j\in\Bbb Z}(J_{n,j}\times I_{n,j}).$$