No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.

- $W$ is a standard Brownian motion.
- $U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$.

Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$.

Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov.

As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.