The answers to Questions (2) and (3) are positive, and the answer to Question (1) is negative. The positive answers follow by a version of the "Lefschetz principle", often also called "spreading out".
Let $X\subset \mathbb{P}^n_{\mathbb{C}}$ be a closed subscheme. Denote by $\mathcal{I}$ the corresponding ideal sheaf. By ampleness / Serre vanishing, there exists an integer $d\geq 0$ such that $\mathcal{I}(d)$ is globally generated. By finiteness of cohomology of coherent sheaves on proper schemes, the $\mathbb{C}$-vector space $H^0(\mathbb{P}_{\mathbb{C}}^n,\mathcal{I}(d))$ has finite dimension.
Let $f_1,\dots,f_m$ be a basis for this finite-dimensional $\mathbb{C}$-vector space. For a choice of homogeneous coordinates, i.e., for a basis $(x_0,\dots,x_n)$ for $H^0(\mathbb{P}^n_{\mathbb{C}},\mathcal{O}(1))$, each $f_i$ is a linear combination of monomials, $$f_i =\sum_{(e_0,\dots,e_n)} c_{i;e_0,\dots,e_n} x_0^{e_0}\cdots x_n^{e_n}, $$ where the indexing set is all $(n+1)$-tuples of nonnegative integers that sum to $d$.
Let $R\subset \mathbb{C}$ be the finitely generated $\mathbb{Z}$-algebra generated by all of the coefficients $c_{i;e_0,\dots,e_n}$. This is an integral domain that contains $\mathbb{Z}$. Since the coefficients are in $R$, every $f_i$ is a well-defined element in $H^0(\mathbb{P}^n_R,\mathcal{O}(d))$. Let $X_R$ denote the zero scheme of $f_1,\dots,f_m$ inside $\mathbb{P}^n_R$. Then $X_R$ is a finite type scheme over $\text{Spec}\ R$. The base change of $X_R$ by $R\hookrightarrow \mathbb{C}$ equals $X$.
By Grothendieck's Generic Freeness result, after replacing $R$ by a dense, Zariski open affine subscheme, the $R$-scheme $X_R$ is $R$-flat. Since the base change to $\mathbb{C}$ is smooth, after shrinking further, also $X_R$ is $R$-smooth.
Since $R$ is a finitely generated integral domain over $\mathbb{Z}$ that contains $\mathbb{Z}$, also $R\otimes_{\mathbb{Z}}\mathbb{Q}$ is a finitely generated $\mathbb{Q}$-algebra that is an integral domain that contains $\mathbb{Q}$. By Zorn's Lemma, there exists a maximal ideal $\mathfrak{m}$ in this nonzero ring. By the Nullstellensatz, for every maximal ideal $\mathfrak{m}\subset R\otimes_{\mathbb{Z}}\mathbb{Q},$ the corresponding residue field $K$ is a finite field extension of $\mathbb{Q}$. Thus, there exists a finite field extension $K/\mathbb{Q}$ and there exists a homomorphism of commutative rings with $1$, $f:R\to K$. Denote by $X_K$ the base change $K$-scheme, $$X_K = X_R\times_{\text{Spec}\ R} \text{Spec}\ \mathbb{C}.$$
Now define $R_{\mathbb{C}}$ to be $R\otimes_{\mathbb{Z}} \mathbb{C}$, define $S$ to be $\text{Spec}\ R_{\mathbb{C}}$, and define $\mathcal{X}\to S$ to be the base change of $X_R\to \text{Spec}\ R$ by $S\to \text{Spec}\ R$. The ring homomorphism $f$ induces a ring homomorphism, $$f_{\mathbb{C}}:R\otimes_{\mathbb{Z}} \mathbb{C} \to K\otimes_{\mathbb{Z}} \mathbb{C}.$$ By the Fundamental Theorem of Algebra, $K\otimes_{\mathbb{Z}}\mathbb{C}$ is a direct product of finitely many copies of $\mathbb{C}$. For any of these factors, say $$p:K\otimes_{\mathbb{Z}}\mathbb{C} \twoheadrightarrow \mathbb{C},$$ the composite $p\circ f_{\mathbb{C}}$ gives a $\mathbb{C}$-point of $S$ such that the fiber of $\mathcal{X}$ over that point equals the base change of $X_K$ by the field homomorphism, $$K\hookrightarrow K\otimes_{\mathbb{Z}}\mathbb{C} \xrightarrow{p} \mathbb{C}.$$ This gives a positive answer to Question (2) and to Question (3).
The answer to Question (1) is negative. There are Abelian varieties $A$ of dimension $2$ defined over a number field $K$ that have a finite set of endomorphisms that do not (simultaneously) admit any infinitesimal deformations to nearby Abelian varieties. The blowing up $X_K$ of $A\times_{\text{Spec}\ K}A$ along the graphs of these endomorphisms gives a $K$-scheme with no nontrivial infinitesimal deformations. So if $X_K$ is not isomorphic to the base change of $\mathbb{Q}$-scheme, then there is also no family $\mathcal{X}\to S$ as in the question.
