**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 $NP^{NP} = NP$.)