By a quantifier free arithmetical sentence I mean a fully quantified sentence in the language of Peano arithemtic $``PA"$, in prenex normal form having no existential quantifier.
By a strict arithmetical sentence I mean a quantifier free arithmetical sentence that doesn't contain any logical connective.
Is PA complete for quantifier free arithmetical sentences?
Formally: do we have: if $P$ is a quantifier free arithmetical sentence, then: $(PA \vdash P) \oplus (PA \vdash \neg P)$?
Is PA complete for strict arithmetical sentences?
Formally: do we have: if $P$ is a strict arithmetical sentence, then: $(PA \vdash P) \oplus (PA \vdash \neg P)$?
First, a comment on terminology: "quantifier free" means, well, "quantifier free," and "arithmetic(al)" is generally understood to refer to arbitrary applications of number quantifiers. Your "quantifier-free" sentences are just universal sentences, and I'll refer to them as such. Meanwhile, your "strict quantifier-free" sentences are just universally quantified equations.
As to your questions, the answers are no and yes(ish) respectively, the crucial point being whether negation is allowed. If negation is allowed, then just note that saying that a Diophantine equation has no solutions is a universal sentence, and we can encode arbitrary $\Pi_1$ queries in this way.
(A bit more explicitly: there is a single Diophantine equation $P(x_1,...,x_n)=0$ such that PA proves $Con(PA)\iff \forall x_1,...,x_n(\neg (P(x_1,...,x_n)=0))$. But the right hand side sentence is universal, since "$P(x_1,...,x_n)=0$" is an atomic formula.)
If however we disallow negation (and don't have "$<$" or "$-$" either - see Andreas' comment below), then we're only looking at sentences of the form "$\forall x_1,...,x_n(P(x_1,...,x_n)=Q(x_1,...,x_n))$" for Diophantine equations $P,Q$. But these sentences are easily seen to be decidable if "$-$" isn't allowed.