Andreas considered the interpretation of your question where we fix the program and then vary the input. Let me now consider the dual version of the question, where we fix the infinite random input and vary the program. Surprisingly, there is something interesting to say.

The concept of asymptotic density provides a natural way to measure the size or density of a collection of Turing machine programs. Given a set $P$ of Turing machine programs, one considers the proportion of all $n$-state programs that are in $P$, as $n$ goes to infinity. This limit, when it exists, is called the asymptotic density or probability of the set $P$, and a set with asymptotic density $1$ will contain more than 99% of all $n$-state programs, when $n$ is large enough, as close to 100% as desired.

What I claim is that for your computational model, almost every program leads to a finite computation.

**Theorem.** For any fixed infinite input (on the read-only tape), the set of Turing machine programs that complete their computation in finitely many steps has asymptotic density $1$.

In other words, for fixed input, almost every program stops in finite time.

The proof follows from the main result of my article: J. D. Hamkins and A. Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47, 2006. http://arxiv.org/abs/math/0504351. The argument depends on the convention in the one-way infinite tape context that computation stops should the head attempt to move off the end of the tape. The idea has also come up on a few other MO quesstions:
What are the limits of non-halting? and Solving NP problems in (usually) polynomial time? in which it is explained that the theme of the result is the black-hole phenomenon in undecidability problems, the phenomenon by which the difficulty of an undecidable or infeasible problem is confined to a very small region, outside of which it is easy.

The main result of our paper is to show that the classical halting problem admits a black hole. In other words, there is a computable procedure to correctly decide almost every instance of the classical halting problem, with asymptotic probability one. The proof method is to observe that on fixed infinite input, a random Turing machine operates something like a random walk, up to the point where it begins to repeat states. And because of Polya's recurrence theorem, it follows that with probability as close as you like to one, the work tape head will return to the starting position and fall off the tape before repeating a state.

My point now is that the same observation applies to your problem. For any particular fixed infinite input, the work tape head will fall off for almost all programs. Thus, almost every program sees only finitely much of the input before stopping.