A theorem in a paper of Peter Heusch, "The Complexity of the Falsifiability Problem for Pure Implicational Formulas" (MFCS'95), seems to suggest the problem is NP-hard. I repeat the first part of its proof here:
By reduction from the restricted version of 3SAT where every variable occurs at most 3 times.
Given such a CNF, WOLOG every variable with 3 occurrences occurs once positively and twice negatively. If $x$ is such a variable, let $C_1$, $C_2$ be the clauses such that $C_1 = \neg x \vee C_1'$ and $C_2 = \neg x \vee C_2'$. Introduce new variables $x'$ and $x''$ and replace $C_1$ and $C_2$ by $\neg x \vee (x' \wedge x'')$, $\neg x' \vee C_1'$, $\neg x'' \vee C_2'$. Repeat.
UPDATED (after the question got changed): All right... your new questions are open questions in complexity theory, as far as I know. There has been some work on derandomizing the Valiant-Vazirani theorem, under reasonable hardness assumptions. A reference:
Adam Klivans, Dieter van Melkebeek: Graph Nonisomorphism Has Subexponential Size Proofs Unless the Polynomial-Time Hierarchy Collapses. SIAM J. Comput. 31(5): 1501-1526 (2002)
So, under some plausible circuit lower bound assumptions, there is a deterministic polynomial time reduction from SAT to USAT. This would give a deterministic reduction from SAT to "Odd-or-Zero-SAT" as well as a deterministic reduction from "Odd-or-Zero-SAT" to USAT.
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(UPDATE: Some stuff got deleted here, as it is no longer relevant to the current version of the question)
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Despite all this, there is an extremely related problem that should be of interest to you. The problem "Parity-SAT" (often written as $\oplus SAT$ in the literature) is the problem of determining whether or not a given Boolean formula has an odd number of assignments. It is well-studied, and is complete for the class $\oplus P$ which contains all languages of the form {$x ~|~ \text{there are an odd number of accepting computation paths in}~N(x)$}, where $N$ is a nondeterministic polynomial time machine.
Now, by the Valiant-Vazirani Theorem (which I suspect you know, since you mentioned USAT) we have
$$SAT ~\leq_R~ \oplus SAT,$$ where $\leq_R$ denotes a randomized polytime reduction. Hence $\oplus SAT$ is "hard" under randomized reductions.
It is not known if $NP = \oplus P$, or $UP = \oplus P$. But, as the Valiant-Vazirani Theorem suggests, you can do a hell of a lot with randomized polynomial time and an oracle for $\oplus P$. We are still figuring out everything you can do. Toda's Theorem tells us that the entire polynomial time hierarchy is in $BPP^{\oplus P}$. It could be that even $PSPACE$ is in $BPP^{\oplus P}$. Another interesting fact due to Papadimitriou and Zachos is that $\oplus P^{\oplus P} = \oplus P$. That is, an oracle for $\oplus P$ is superfluous if you already have the power of $\oplus P$. This follows from the observation that the XOR of a bunch of XORs is still an XOR. (Similarly, $P^{P} = P$, but it is not known or believed that $NP^{NP} = NP$.)
Best Answer
In his infamously short paper "Average-case complete problems," Leonid Levin uses a tiling problem as the master ("first") NP-complete average-case problem (which means he also automatically uses it as a master NP-complete problem).
UPDATE: Contrary to what I was speculating in my answer previously, in his original paper on the Cook-Levin theorem (I found an English translation linked to from the Wikipedia Cook-Levin theorem article), it's not clear whether Levin uses the tiling problem as a master problem. He lists six NP-complete problems (SAT is number 3, and the tiling problem is number 6), but leaves out the proofs, so it's not completely clear which one is the master problem. Very likely, it was SATISFIABILITY, and so Levin found essentially the same proof as Cook.