The implication is not true in general, if I understand you correctly. For example, consider the property "affine" for morphism of schemes. Say This property is fpqc locally on the target.
Say, you have a morphism of schemes $f:X \to Y$ such that for every artinian ring $A$ and morphism $\text{Spec}(A) \to Y$ the base change $f_A : X_A = X \times_Y \text{ Spec}(A) \to \text{Spec}(A)$ is affine. Then $f$ is not an affine morphism, in general. For instance, consider for $f$ the open immersion of the pointed affine plane $\mathbb{A}^2_k -\lbrace0\rbrace\to \mathbb{A}^2_k$$X := \mathbb{A}^2_k -\lbrace0\rbrace\to \mathbb{A}^2_k = :Y$.
Then $\mathbb{A}^2_k$ is not affine! Otherwise, the open immersion $f$ would be an isomorphism as $H^0(X,O_X) = H^0(Y,O_Y)$ using that $X$ is $S_2$.
Thus, the open immersion $f$ is not an affine morphism. However, every fiber is either $\mathbb{A}^2$ or $\mathbb{A}^2 - \lbrace 0 \rbrace$. Both are affine curves. So, $f$ has affine fibers, but is not affine. Morever, if you take any Artinian ring $A$, then for every morphism $\text{Spec}(A) \to Y$ the base change $f_A : X_A \to \text{ Spec}(A)$ has affine source $X_A$. To see this, use that $k := A/Nil(A)$ is a field and that we know already that $X_k$ is affine. The map $X_k \to X_A$ is finite, finitely presented and surjective, since $\text{ Spec}(k) \to \text{Spec}(A)$ is. Thus, $X_A$ is affine by Chevalley's Theorem for affines.
Therefore, you cannot disprove that $f$ is affine by testing with Artin rings.
However, if you impose an additional global condition (e.g. properness), then some properties can be checked over Artin rings as Ramsey pointed out.