Update: just realized that there is a very simple argument. $C([0,1])$ is separable, i.e. has a dense countable set and every continuous functional is determined by its values on this set. Hence $|C^*([0,1])|=\mathbb{R}$.
The previous answer is fine but may be this will be useful.
Answer: $|C^*([0,1])|=\mathbb{R}$.
Riesz representation theorem states that any continuous functional $F$ on $C(0,1)$ has the form $$F(f)=\int\limits_0^1f(x)d\Phi(x), $$ where $\Phi(x)$ is a function of finite total variation and the integral is Riemann–Stieltjes.
Since $\Phi(x)$ has finite total variation we have $$\Phi(x)=\varphi^+(x)-\varphi^-(x) $$ for two monotonic functions on $[0,1]$.
Monotonic function $\varphi$ can have only countable many points of discountinuity. Denote them by $a_1,a_2,...$ and denote by $b_n:=\lim\limits_{x\to a_i+0}\varphi(x)-\lim\limits_{x\to a_i-0}\varphi(x)$. So $$\varphi(x)=\sum\limits_{n=1}^{\infty} b_n \delta(a_i) + \widetilde\varphi(x), $$ where $\delta(a_i)$ is the Dirac delta function at point $a_i$ and $\widetilde\varphi(x)$ is continuous.
So $\Phi(x)$ is determined by the following data: $$a_n^+,b_n^+,\widetilde\varphi^+,a_n^-,b_n^-,\widetilde\varphi^-. $$
So the cardinality of $C^*([0,1])$ is bounded from above by $$\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}=\mathbb{R} $$ (recall that $\widetilde\varphi^\pm$ are continuous, and that continuous function is determined by its values on some countable dense set, so $|C([0,1])|=\mathbb{R}$).