Step 1
Take any first-order structure $M$ over any language $L$.
Then $M$ already interprets formulae over $L$, and we can use the interpretation. For example:
$\forall x,y\ \exists z\ ( z \ne x \land P(x,f(y,z)))$
simply becomes:
$\forall x,y \in D\ \exists z \in D\ ( z \ne x \land P_M(x,f_M(y,z)))$
where $D$ is the domain of $M$ and $P_M,f_M$ are the interpretations of $P,f$ in $M$.
It is trivial to translate the function application and ordered pairs to pure ZFC, but I won't do it since it is not worth the trouble.
Step 2
Now it seems you also want to get rid of even quantifiers.
Take any $1$-parameter sentence $φ$ over ZFC.
Then:
$
\def\none{\varnothing}
$
$\forall x \in D\ ( φ(x) )$ iff $\{ x : x \in D \land φ(x) \} = D$.
$\exists x \in D\ ( φ(x) )$ iff $\{ x : x \in D \land φ(x) \} \ne \none$.
Thus you can recursively get rid of quantifiers, if you allow the use of set constructors, which pure ZFC does not have.
Notes
If you forbid set constructors, then "$\forall x,y \in D\ ( R_M(x,y) \to R_M(y,x) )$" cannot be translated to a sentence without quantifiers, because any translation must assert that $R$ is symmetric, and there is no way to do that without quantifiers. Notice that set equality itself has a quantifier hiding inside, which is crucial in allowing equality between constructed sets to capture quantified assertions.
We define a valuation $v$ on all the prime formulas of the language such that:
$v(\alpha)= \text T \text { iff } U \vDash \alpha[s]$.
Then we extend it to $\overline v$ such that $\overline v(\varphi) = \text T \text { iff } U \vDash \varphi [s]$, for a formula $\varphi$ whatever of the language.
That such a $\overline v$ exists is proved by induction.
The base case is already in the definition for prime formulas. The induction step needs the basic connectives of the language. Example for $\lnot$: assume that the property hold for $\psi$ and let $\varphi := \lnot \psi$.
We have:
$\overline v(\varphi) = \text T \text { iff } \overline v (\lnot \psi)= \text T \text { iff } \overline v (\psi)= \text F \text { iff } U \nvDash \psi[s] \text { iff } U \vDash \lnot \psi [s] \text { iff } U \vDash \varphi[s].$
Similar for $\to$ and $\lor$, etc. A little bit more tricky the case for $\forall$.
Now for the result: assume that $\Gamma \vDash_{\text {Taut}} \varphi$ and $\Gamma \nvDash \varphi$.
This means that, for some $U$ and $s$ we have:
$U \vDash \gamma_i[s]$, for every $\gamma_i \in \Gamma$, and $U \nvDash \varphi [s]$.
But this means that, using the construction above, we can find $v$ such that $\overline v(\gamma_i)= \text T$ and $\overline v (\varphi)= \text F$, contradicting the assumption that $\Gamma \vDash_{\text {Taut}} \varphi$.
Best Answer
Yes.
Every formula that is a first-order instance of a propositional tautology, i.e. one that can one get by substituting first-order formulas for propositional letters in a propositionally valid formula, is also valid in FOL.