Collatz-like properties of finite fields

Here is a proof of the generalization of your Weak conjecture to the ring $\mathbf{Z}/m\mathbf{Z}$ where $m$ is any odd positive integer. First let me clarify what is being proved. Let $S$ be the subgroup of $(\mathbf{Z}/m\mathbf{Z})^\times$ which is generated by $2$, and consider a directed graph whose vertices are the sets $gS$ with $g\in\mathbf{Z}/m\mathbf{Z}$ (I emphasize that $g$ need not be in $(\mathbf{Z}/m\mathbf{Z})^\times$), and which has a directed edge from $gS$ to $(3g+1)S$ for every $g\in\mathbf{Z}/m\mathbf{Z}$. I will show:

1) If $3\nmid m$ then this directed graph is strongly connected (in the sense that there is a directed path from any vertex to any other vertex).

2) If $3\mid m$ then for any $g,h\in(\mathbf{Z}/m\mathbf{Z})^\times$ there is a directed path from the vertex $gS$ to the vertex $hS$ (although this path might go through vertices of the form $iS$ with $i\notin (\mathbf{Z}/m\mathbf{Z})^\times$).

Both of these are special cases of the Theorem proved below.

Lemma. Let $n$ be a positive integer, and let $M$ be a subset of $\mathbf{Z}/n\mathbf{Z}$ which contains elements $u,v$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Then, for every sufficiently large positive integer $s$, every element of $\mathbf{Z}/n\mathbf{Z}$ is the sum of exactly $s$ elements of $M$.

Proof. Let $M_r$ be the set of all sums of exactly $r$ elements of $M$, so that $M_1=M$. Pick $u,v\in M_r$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Plainly $M_{r+1}$ contains $M_r+u$ and $M_r+v$, so that $\#M_{r+1}\ge\#M_r$. If $\#M_{r+1}=\#M_r$ then we must have $M_r+u=M_r+v$, or equivalently $M_r+(u-v)=M_r$. But then $M_r+\ell(u-v)=M_r$ for any positive integer $\ell$, whence $M_r+w=M_r$ for any $w\in\mathbf{Z}/n\mathbf{Z}$, so we must have $M_r=\mathbf{Z}/n\mathbf{Z}$. Thus, for every $r>0$, either $\#M_{r+1}>\#M_r$ or $M_r=\mathbf{Z}/n\mathbf{Z}$. Since $\#M_r\le m$ for every $r$, it follows that there is some $r$ for which $M_r=\mathbf{Z}/n\mathbf{Z}$, whence $M_s=\mathbf{Z}/n\mathbf{Z}$ for every $s\ge r$. This finishes the proof of the Lemma.

Theorem. Write $m=3^kn$ where $k\ge 0$ and $3\nmid n$. Pick any $g,h\in\mathbf{Z}/m\mathbf{Z}$, and suppose that either $k=0$ or $3\nmid h$. Then there is a directed path from $g\langle 2\rangle$ to $h\langle 2\rangle$.

Proof. Apply the Lemma with $M$ being the subgroup of $(\mathbf{Z}/n\mathbf{Z})^\times$ generated by $2$ (and with $u=2$ and $v=1$) to obtain a corresponding $s$. Let $r$ be the order of $3$ in $(\mathbf{Z}/n\mathbf{Z})^\times$. Let $\ell$ be the smallest integer such that $\ell>k/r$. Recall that every element of $U:=(\mathbf{Z}/3^k\mathbf{Z})^\times$ is a power of $2$. Since an element $i\in \mathbf{Z}/3^k\mathbf{Z}$ lies in $U$ if and only if $i+3/2$ lies in $U$, it follows that every element of $U$ can be written as $-3/2+2^t$ for some positive integer $t$. The hypothesis that either $k=0$ or $3\nmid h$ implies that the image of $h$ in $\mathbf{Z}/3^k\mathbf{Z}$ is actually in $U$, so there is some $t>0$ for which $h\equiv -3/2+2^t\pmod{3^k}$. Let $b_0,\dots,b_{s-1}$ be nonnegative integers for which $h-3g+s-2^t\equiv\sum_{i=0}^{s-1} 2^{b_i}\pmod{n}$. Define $$a_0:=t,$$ $$a_i:=0 \,\,\text{if either $0<i<r\ell$ or $r\nmid i$},$$ $$a_{r(\ell+j)}:=b_j \,\,\text{for $0\le j\le s-1$}.$$ Then $$x:=3^{r(\ell+s-1)+1}2^0g+\sum_{i=0}^{r(\ell+s-1)} 3^i 2^{a_i}$$ equals $$3^{r(\ell+s-1)+1}g + \sum_{i=0}^{r(\ell+s-1)} 3^i + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1)$$ $$=3^{r(\ell+s-1)+1}g + \frac{3^{r(\ell+s-1)+1}-1}{3-1} + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1).$$ Here $x$ is congruent mod $3^k$ to $$-\frac12+(2^t-1)=2^t-\frac32=h.$$ Next, $x$ is congruent mod $n$ to $$3g+\frac{3-1}{3-1}+\sum_{j=0}^{s-1} (2^{b_j}-1) + (2^t-1)= 3g+1+(h-3g+s-2^t)-s+(2^t-1)=h.$$ Therefore $x=h$. An easy induction shows that the union of the sets $v\langle 2\rangle$ for which the vertex $v\langle 2\rangle$ can be reached from $g\langle 2\rangle$ via a directed path of length $z$ is $\{3^z2^{d_z}g +\sum_{i=0}^{z-1} 3^i 2^{d_i}\colon d_i\in\mathbf{Z}\}$. Thus there is a directed path of length $r(\ell+s-1)+1$ from $g\langle 2\rangle$ to $h\langle 2\rangle$. This finishes the proof of the Theorem.

Shifting by $1/2$, one sees that the map $x\mapsto 3x+1$ is conjugate to multiplication by 3. It is periodic with period the multiplicative order of 3. Therefore, strong conjecture cannot hold if this order is strictly less than the number of cosets of $\langle 2\rangle$ --- there are just too many cosets to visit. This presumably explains your counterexamples with "unusually high" number of cosets.