As Piotr pointed out, the answer to your question is no as stated.
I just wanted to add a further comment (too long for an actual comment)
I guess what you're thinking is that it suffices to check etaleness at closed points, which is true.
However, it would be wrong to assume that all closed points of $X$ should lie on the special fiber. This is true if $X\rightarrow\text{Spec }R$ is proper, but false in general. This is the key point in Piotr's counterexample. The ramification locus of his map $f : X = \mathbb{A}^1_R\rightarrow\mathbb{A}^1_R = Y$ given by $f(x) = x+tx^n$ is a union of closed points which do not lie on the special fiber. If you consider his map, and extend it naturally to a map $\overline{X} = \mathbb{P}^1_R\rightarrow\mathbb{P}^1_R = \overline{Y}$, there you would be able to detect the ramification locus on the special fiber of $\overline{X}$. Indeed, the ramification locus is the divisor determined by the $(n-1)$th roots of $\frac{-1}{nt}$ in $\text{Frac}(R)$ (viewed as points on the generic fiber, the corresponding divisor being their closure in $X$). These elements of $\text{Frac}(R)$ are not integral over $R$, so the corresponding divisor intersects the special fiber of $\overline{X}$ at $\infty$. However, when you restrict back to $X = \mathbb{A}^1_R$, the divisor no longer intersects the special fiber. Indeed, since taking the closure of the $(n-1)$th roots of $\frac{-1}{nt}$ inside $\overline{X}$ only adds the $\infty$ of the special fiber, upon restricting to $X$, the ramification divisor becomes a disjoint union of closed points, all lying on the generic fiber.
This is also an interesting example of a nice (regular Noetherian) scheme of dimension 2 for which not all closed points have local rings of the same dimension. I believe this failure is stated as saying "the scheme is not equicodimensional" (it seems to be rather difficult to find references for this, but IIRC once upon a time I found a discussion of it in EGA...somewhere). For this, you can even take the scheme $\text{Spec }\mathbb{Z}_{(p)}[x]$.