Here's a tentative answer to the first question. I'd be glad for any comment, and hints regarding the second question!

Let $B \xleftarrow{p} A \xrightarrow{q} C$ be a weak pullback of
$B \xrightarrow{f} D \xleftarrow{g} C$, and consider any span
$B \xleftarrow{u} G \xrightarrow{v} C$ such that $fu = gv$.

Consider the pullback $B \xleftarrow{\pi} P \xrightarrow{\pi'} C$ of $f$ and $g$, which is constructed just as in graphs, with identities given by pairs of identities.

For any edge $e \colon x \to x'$ in $G$, we construct an identity-free path $p_e \colon (u(x),v(x)) \to^\star (u(y),v(y))$ in $P$, by induction on the lengths of $u(e)$ and $v(e)$ (which are identity-free paths in $B$ and $C$, respectively):

if one is the empty path, w.l.o.g. $u(e)$ is the empty path on, say, $B_0$, then the other, say $C_0 \xrightarrow{c_1} C_1 \ldots \xrightarrow{c_n} C_n$, must map to an identity-only path in $D$, and we let $e_p$ consist of all pairs $(id_{B_0}, c_i)$, which do form a path in $P$;

otherwise, we consider the first edges $b_1$ and $c_1$:

if none of them maps to an identity (by $f$ resp. $g$), then we pick $(b_1,c_1)$ as the first edge of $e_p$ and continue by induction;

if at least one of them, w.l.o.g. $b_1$, maps to an identity, then we pick $(b_1, id)$ and continue by induction (note that in this case we do not decrease the length of the image path in $D$).

Let $l_e$ be the length of $p_e$.

Now, let $G'$ denote the free reflexive graph with

as vertices those of $G$, plus $l_e - 1$ vertices $x^e_1, \ldots, x^e_{l_e - 1}$ for each edge $e$ of $G$,

for each edge $e \colon x \to x'$ of $G$, a path $x \to x^e_1 \to \ldots \to x^e_{l_e - 1} \to x'$ (i.e., really $l_e$ edges).

Our paths $e_p$ determine a morphism $k \colon G' \to P$ of reflexive graphs, such that $\pi k$ and $\pi' k$ coincide with $u$ and $v$ on vertices, and $f \pi k = g \pi' k$.

Thus by universal property of weak pullback, we obtain a morphism $h \colon G' \to A$ such that $p h = \pi k$ and $q h = \pi' k$.

This yields a morphism $M(h) \colon M(G') \to M(A)$ such that
$M(p) M(h)$ and $M(q) M(h)$ respectively map each $p_e$, viewed as an edge in $M(G')$, to $u(e)$ and $v(e)$. Precomposing with the morphism $G \to M(G')$ mapping each $e$ to $p_e$, we obtain the desired morphism $j \colon G \to M(A)$ such that $M(p) (j(e)) = u(e)$ and $M(q) (j(e)) = v(e)$.