Consider a transfinite Turing Machine model, allowing ordinal-length tape and time, and a number of states of arbitrary cardinality.
Is there an algorithm P with at most $\aleph_{0}$ states, such that P halt after $\omega_{1}$ steps on the empty input?
The answer is no, there is no such program.
Let me say, first, that allowing countably infinitely many states into the program is basically equivalent to having a finite program with a real parameter, since one can have a master control program that consults the real parameter to see what it should do next; this would align things a little closer with how the theory is usually undertaken. But it is of no consequence, and I can go along with your infinite program.
Suppose that $p$ is one of your programs, and we run it on an ordinal-length empty tape, and suppose that it halts at time $\omega_1$. Consider the relativized constructible universe $L[p]$, and observe that the operation of the program $p$ is absolute to this universe $L[p]$. Indeed, if $\theta>\omega_1$, then the operation of the machine is absolute to $L_\theta[p]$, which will therefore observe that the program halts at stage $\omega_1$. By the Löwenheim-Skolem theorem and condensation, there is a countable ordinal $\gamma$ with an elementary embedding $L_\gamma[p]\precsim L_\theta[p]$. It follows that $L_\gamma[p]$ also observes that $p$ halts at the ordinal $\omega_1^{L_\gamma[p]}$, which is a countable ordinal less than $\gamma$. But the operation of the machine at stages below $\gamma$ is absolute to $L_\gamma[p]$, and so the program must really halt at that countable stage, contradicting out assumption.
The argument shows that every such program, even allowing countably infinitely many states, must halt at a countable stage, if it halts at all.
Another way to see this is to point out that the halting nature of a program is a $\Sigma_1$ expressible fact about the program, and so if this $\Sigma_1$ statement about $p$ is true, then it will become true by the first $\Sigma_1(p)$ stable ordinal, which is countable.
Meanwhile, the very first theorem of my paper
shows that in the case of infinite time Turing machines, which allow a tape of length $\omega$ rather than Ord as in your question, every halting computation must halt at a countable stage. Indeed, every computation either halts or strongly repeats at a countable stage.
