I will answer the question in a slightly more specific context, where

- We restrict ourselves to the projective plane $\mathbb{P}_2(R)$ over $R$, i.e. triples $(x,y,z)\in R^3$ such that $xR+yR+zR=R$, up to multiples in $R^\ast$.
- We use the given Hermitian form $h\big((x_1,y_1,z_1),(x_2,y_2,z_2)\big) = x_1z_2^\theta+z_1x_2^\theta+y_1y_2^\theta$

If we have a triple $v=(x,y,z)$ in the projective plane satisfying $h(v,v)=0$, that implies that either $x$ or $z$ is invertible, so we may assume $z$ is invertible, and hence we can assume $z=1$.

In this formulation, the question reduces to the following:

If $x+x^\theta+yy^\theta\in M$, can we find $x',y'$ such that $x'+x'^\theta+y'y'^\theta=0$ and $x-x',y-y'\in M^i$.

Define for any ideal $I$ in $R$ for which $I^\theta=I$
$$S_I = \{x+x^\theta+yy^\theta\mid x,y\in I\}.$$
**Property**: The following two statements are equivalent for $I$ an ideal in $R$ (with $I^\theta=I$):

- $S_I = S_R\cap I$
- If we have a solution $x+x^\theta+y^{1+\theta}\in I$, then we can find $x',y'$ such that $x'+x'^\theta+y'^{1+\theta}=0$ and $x-x',y-y'\in I$.

*Proof*: Assume that $S_I = S_R\cap I$, and that we have $x+x^\theta+y^{1+\theta}= i\in I$, then by the assumption there are $s,t\in I$ such that $s+s^\theta+t^{1+\theta}=i$. We can take $x' = x-s+y^\theta t-t^{1+\theta}$ and $y' = y-t$ and get
$$x'+x'^\theta+y'^{1+\theta} = (x+x^\theta+y^{1+\theta})-(s+s^\theta+t^{1+\theta}) = 0\;.$$
Clearly, $x-x'$ and $y-y'$ are in $I$.

Next, assume 2. holds. The inclusion $S_I\subseteq S_R\cap I$ is immediate, so we show the other inclusion. Assume we have $i\in S_R\cap I$, i.e. we have $x,y\in R$ such that $x+x^\theta+y^{1+\theta}= i$. By the assumption, we can find $x',y'$ such that $x'+x'^\theta+y'^{1+\theta}=0$ and $x-x',y-y'\in I$. We can now set $s = x-x'+y'^\theta(y-y')$ and $t = y-y'$. Then
$$s+s^\theta+t^{1+\theta} = x+x^\theta+y^{1+\theta}=i$$
and $s,t\in I$. Hence $i\in S_I$.

This seemingly easy reformulation does give an easier check, as one can usually compute $S\cap I$ and $S_I$ if a ring and involution are given.

**Corollary**: If the residue field $R/M$ has characteristic different from $2$, the answer to question 1 is yes.

*Proof*: We apply the previous with as the ring $R/M^i$ and as ideal $M$. If $x+x^\theta+y^{1+\theta}= m \in M$, then $m=m^\theta$, so we can take $s=m/2$ and $t=0$ to get $s+s^\theta+t^{1+\theta}=m$ with $s,t\in M$. Hence $S_{R/M^i}\cap M \subseteq S_M$, so we get an affirmative answer to question 1.

Lastly, we now know that to get a counterexample, we need to look at local rings for which the residue field has characteristic $2$. One can check that if we take $R = \mathbb{Z}_2$ with trivial involution $\theta$, that $(1,0,1)$, a solution in $\mathbb{Z}_2/2\mathbb{Z}_2$, is never reached as a projection from a solution in $\mathbb{Z}_2/2^k\mathbb{Z}_2$ for $k\geq2$.

Credit for the equivalency goes to one of my colleagues, who is not active on MO.