I stumbled onto a proof that this is true assuming the Collatz iteration always ends in a finite cycle (so no values escape to infinity). It's a little tricky, but very short, so any mistakes will turn up quick.
I'll write $\rightarrow$ for a standard iteration and $\implies$ for a possibly non-standard one. A number that fails to satisfy the conjecture because it can't get to $1$ I'll call an escapee.
First of all, a lemma: for all $x$, $4 x \implies 12 x + 1 \rightarrow 36x + 4 \rightarrow 18x + 2 \rightarrow 9 x + 1$.
Now let $x$ be the first escapee. By elementary considerations, $x \equiv 7 \pmod{12}$, and $4x$ is also an escapee because we can cycle back to it from $x$.
Suppose $x = 24k + 7$. Then we have:
$4(24k+7) \implies 9(24k+7)+1 \rightarrow 108k+32 \rightarrow 54k+16$.
But also, $18k+5 \rightarrow 54k+16$. This means $18k+5$ is an escapee too, but it can't be, since it's less than $x$. So the hypothesis is false and we have $x = 24k + 19$ instead.
Now consider:
$4(24k+19) \implies 216k + 172 \rightarrow 108k + 86 \rightarrow 54k + 43 \rightarrow 162k + 130 \rightarrow 81k+65$.
By similar reasoning, $k \neq 3 \pmod{4}$, or else we can continue the chain to get an escapee less than $x$. Finally,
$24k+19 \rightarrow 72k+58 \rightarrow 36k + 29 \rightarrow 108k + 88 \rightarrow 54k + 44 \rightarrow 27k + 22$
shows that $k$ can't be even, and substituting $k = 4 j + 1$:
$27k + 22 = 108j + 49 \rightarrow 324j + 148 -> 162j + 74 \rightarrow 81j + 37$
completes the proof, since that's all the cases.