Platonic Truth and 1st Order Predicate Logic
The phenomenon accords more strongly with your philosophical explanation if you ask also that the sentences have complexity $\Pi^0_1$. That is, the universal statement $\forall x\ \varphi(x)$ should have $\varphi(x)$ involving only bounded quantifiers, so that we can check $\varphi(x)$ for any particular $x$ in finite time. If you drop that requirement, there are some easy counterexamples, hinted at or given already in the comments and other answers.
But meanwhile, even in the case you require $\varphi(x)$ to have only bounded quantifiers, there is still a counterexample.
Theorem. If $\newcommand\PA{\text{PA}}\newcommand\Con{\text{Con}}\PA$ is consistent, then there is a consistent theory $T$ extending $\PA$ with two $\Pi^0_1$ sentences $$\forall x\ \varphi(x)$$ $$\forall x\ \psi(x)$$ both of which are consistent with and independent of $T$, but which are not jointly consistent with $T$.
Proof. Let $T$ be the theory $\PA+\neg\Con(\PA)$, which is consistent if $\PA$ is consistent. Let $\rho$ be the Rosser sentence of this theory, which asserts that the first proof in $T$ of $\rho$ comes only after the first proof of $\neg\rho$ (see also my discussion of the Rosser tree). Our two $\Pi^0_1$ sentences are:
- every proof of $\rho$ from $T$ has a smaller proof of $\neg\rho$.
- every proof of $\neg\rho$ from $T$ has a smaller proof in $T$ of $\rho$.
The first statement is equivalent to $\rho$, and the second is equivalent over $T$ to $\neg\rho$, since $T$ proves that every statement is provable; the only question is which proof comes first. So both statements are consistent with $T$.
But the sentences are not jointly consistent with $T$, since in any model of $T$, both $\rho$ and $\neg\rho$ are provable from $T$, and so one of the proofs has to come first. QED
This post is not an answer to your question, but it explains the reason that if $\bf GC$ (or $\bf RH$) is independent of $\bf ZFC$, $\bf GC$ (or $\bf RH$) is true in the standard model of natural numbers.
The reason is $\bf GC$ and $\bf RH$ are $\Pi_1$ sentences in the language of arithmetic.
Def.
- $x|y := \exists z(z \leq y \land x\cdot z = y)$
- $Pr(x) := \forall y(y<x\land y>0 \land y|x \to y=1)$
Therefor $\bf GC$ can be defined by $\forall x\exists y,z(y+z = 2\cdot x+4 \land Pr(y) \land \Pr(z))$ which is a $\Pi_1$ sentence. For $\Pi_1$ definition of $\bf RH$ see here.
Let $\phi(x)$ be a $\Delta_0$ formula and suppose ${\bf PA} \nvdash \exists x \neg \phi(x)$, then $\mathbb{N}\models \forall x \phi(x)$. This is true because of $\Sigma_1$ complentness of ${\bf PA}$, that is if $\psi$ be a $\Sigma_1$ sentence, then $N\models \psi$ iff ${\bf PA}\vdash \psi$.
The important thing in this argument is $\Pi_1$ definability of problem. For example consistency of $\bf PA$ is a $\Pi_1$ sentence and by second incompleteness theorem, ${\bf PA} \nvdash Con_{\bf PA}$, but $\mathbb{N}\not\models \neg Con_{\bf PA}$, therefore we can not prove similar theorems for formulas in another level of Arithmetical Hierarchy except $\Pi_1$.
It cannot be done in a consistent way.
Consider a closed statement $\psi$ which is independent of a theory $T$, and take $\forall x . \psi$ and $\forall x . \lnot\psi$. (I made the closed statement have a dummy free variable to satisfy your condition.) Both statements are of the kind you are asking for, but when we add both to $T$ we get an inconsistent theory.
It should be clear that one can come up with examples where the two sentences that contradict each other are not so blatantly in opposition with each other. And with a bit of work we can even come up with examples where the free variable $x$ is doing something.