Hi lpdbw,
I think this is a very interesting questions, here at least a partial answer; It depends heavily on the little word "strong" in parentheseis.
Assuming that $(X_t)$ is strong Markov, the answer to your questions seems to be "yes": The completed filtration of an $\mathbb{R}^d$-valued strong Markov process is already right-continuous (cf. Theorem 2.7.7. of Karatzas/Shreve: Brownian Motion and Stochastic Calculus - note that they use a different terminology for augmented/completed). And just augmenting the filtration does not affect the martingale property (these are just sets of mass zero).
If you look however on Markov processes which does not necessarily has the strong Markov property, the answer seems to be negative as seen in the following counterexample: Let $(W_t)$ be a Brownian motion in its natural filtration and $\xi$ a random variable independent of the Brownian filtration, $\mathbb{P}[\xi =1] = \mathbb{P}[\xi =2] =\frac{1}{2}$, and define the process $(X_t)$ as
$$
X_t = \int_0^t \xi \; \;dW_s.
$$
This process is a Markov process and a martingale in it's natural filtration $\mathcal{F}_{t}$, but just passing to the right continuous filtration $\mathcal{F}_{t+} = \bigcap_{s>t}\mathcal{F}_s$ destroys the Markov property. Note that for $t>0$ we have $\mathcal{F}_t =\mathcal{F}_{t+} = \sigma(W_s; s\leq t) \vee \sigma(\xi)$, however at $t=0$ they are fundamentally different: $\mathcal{F}_0$ is trivial whereas $\mathcal{F}_{0+} = \sigma(\xi)$. Adding additional null sets changes again nothing.
It is true that $\langle M\rangle^n_1 \to \langle M\rangle_1$ in $L^1$ when $M$ is a continuous square-integrable martingale. In fact, the result holds even if $M$ is cadlag, as long as $\langle M\rangle$ is continuous.
Indeed, $M^2$ is then a submartingale of class D and so, since
$$E[(M_{t_{i+1}} - M_{t_{i}})^2 |\mathcal{F}_{t_i}]=E[M_{t_{i+1}}^2 - M_{t_{i}}^2 |\mathcal{F}_{t_i}],$$
the result follows from the analogous results for the predictable component of the Doob decomposition of a submartingale of class D, which was proved in
Doléans, Catherine. "Existence du processus croissant naturel associé à un potentiel de la classe (D)." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 9, no. 4 (1968): 309-314.
and which constitutes Theorem 31.2, chapter 6 of
Rogers, L. Chris G., and David Williams. Diffusions, Markov processes and martingales: Volume 2, Itô calculus. Vol. 2. Cambridge university press, 2000.
When $M$ is cadlag (and $\langle M\rangle$ is not continuous), $\langle M\rangle^n_1$ converges to $\langle M\rangle_1$, but only in the $\sigma\left(L^{1}, L^{\infty}\right)$-topology, see for example
Rao, K. Murali. "On decomposition theorems of Meyer." Mathematica Scandinavica 24, no. 1 (1969): 66-78.
In this case, it follows from Mazur's lemma that one can obtain (strong) convergence in $L^1$ by
replacing $(\langle M\rangle^n_1)_n$ with a forward convex combinations thereof. In fact, taking forward convex combinations, one can obtain convergence in $L^1$ simultaneously at all times $t\in [0,1]$, as I showed in
Siorpaes, Pietro. "On a dyadic approximation of predictable processes of finite variation." Electronic Communications in Probability 19 (2014): 1-12.
Best Answer
No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.
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.