I stumbled onto a proof that this is true assuming the Collatz iteration always ends in a finite cycle (no values escape to infinity).  It's very short, so any mistakes will turn up quick.

I'll write $\longrightarrow$ for a standard iteration and $\implies$ for some possibly non-standard iterations.  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 $x$, $4 x \implies 9 x + 1$.

Proof: $4x \implies 12 x + 1 \longrightarrow 36x + 4 \longrightarrow 18x + 2 \longrightarrow 9 x + 1$.

Now let $x$ be the first escapee.  By elementary considerations, $x \equiv 7 \pmod{12}$, and $4x$ is also an escapee because we're assuming that we can cycle back to it from $x$.  (And this is the only time we'll need that assumption, so if for some reason $x \implies 4x$ anyway, that would be good enough to make the result unconditional.)

Suppose $x = 24k + 7$.  Then we have:

$4x = 4(24k+7) \implies 9(24k+7)+1 = 216k+64 \longrightarrow 108k+32 \longrightarrow 54k+16$.

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:

$4x = 4(24k+19) \implies 216k + 172 \longrightarrow 108k + 86 \longrightarrow 54k + 43 \longrightarrow 162k + 130 \longrightarrow 81k+65$.

By similar reasoning, $k \neq 3 \pmod{4}$, or else we can continue the chain two more steps to get an escapee less than $x$.  Finally,

$x = 24k+19 \longrightarrow 72k+58 \longrightarrow  36k + 29 \longrightarrow 108k + 88 \longrightarrow 54k + 44 \longrightarrow 27k + 22$.

shows that $k$ can't be even, and substituting $k = 4 j + 1$:

$27k + 22 = 108j + 49 \longrightarrow 324j + 148 \longrightarrow 162j + 74 \longrightarrow 81j + 37 < x$

completes the proof, since that's all the cases.