Show that the halting problem is decidable for one-pass Turing machines

$L=\{<\!M,x\!>\, \mid M's \text{ transition function can only move right and } M\text{ halts on } x \}$. I need to show that $L$ is recursive/decidable.

I thought of checking the encoding of $M$ first and determine whether its transition function moves only right (Can I do that?). If so then try to simulate $M$ on $x$ for $|Q|+1$ steps, if it stops then $<\!M,x\!>\, \in L$ otherwise it is not.

Is this correct?

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By R do you mean Regular Languages? –  Dave Clarke Jun 3 '12 at 14:53
No, to recursive languages. Thank you for the clarification. –  Numerator Jun 3 '12 at 14:58
See this question. –  Dave Clarke Jun 3 '12 at 15:02

I assume $Q$ is state set of $M$. If so, this runtime bound does not make much sense; in particular, a Turing machine with three states can move to the end of the input and check whether $x$ is an even number; it is clearly not enough to wait three steps.
The next question is whether there is a computable bound on the runtime of halting one-pass machines. Unfortunately, there is not: $M$ might be nondeterministic and halt after an arbitrary amount of steps.
So we determinise $M$ while simulating (that is computable) and look for a bound for all paths. Now we are getting somewhere: after having consumed the input, there are two cases. Either $M$ halts or it enters a loop. As it only moves right (into empty tape), such loops can be detected.
Putting things together, we simulate a determinisation of $M$. We run every branch until $x$ is consumed¹ and then for another $|Q|\cdot|\Sigma_T|$ steps, $Q$ the set of states and $\Sigma_T$ the tape alphabet of $M$. At this point $M$ has either stopped or loops (in this branch), which we detect by a pair of current state and tape symbol occuring for the second time. We only have to check finitely many branches up to a computable bound, therefore $L$ is recursive.
1. $M$ might loop here already if not moving the head is allowed, but that can be detected, too.