Does $R$ a domain imply $\operatorname{gr}(R)$ is a domain?

No. Let $R=\mathbb{C}[x,y]/(y^2-x^3-x^2)$ be the coordinate ring of a singular cubic. Filter $R$ by total degree in $x$ and $y$. Check that $R$ is a domain (i.e., the cubic is irreducible) but that $(x-y)(x+y)=x^3=0$ in the associated graded ring.

What is going on: analytically locally, the curve is reducible near $0$: there are branches approximated by $y-x$ and $y+x$. So the completion wrt the filtration above is not a domain, and hence its associated graded, which is the same as the associated graded of the original ring, cannot be a domain.

A slightly simpler example: find a filtration on $k[x,y,z]/(xy-z^2)$ such that the associated graded ring is $k[x,y,z]/(xy)$.

The answer is no. Here is a counter-example:

Let $R = \mathbb C[x,y]/(y^2 - x^3 -x^2),$ with the filtration being the one induced by the filtration by degree on $\mathbb C[x,y]$ (i.e. take the quotient filtration on $R$). This is the same as the $\mathfrak m$-adic filtration on $R$, where $\mathfrak m$ is the ideal generated by $x$ and $y$.

The $R$ is a domain, but $gr(R)$ is not: $y^2 - x^2 = (y - x)(y+x) = x^3$, and so $y-x$ and $y + x$ are non-zero elements of $gr^1$ whose product vanishes (because the element $x^3$ has vanishing image in $gr^2$).

Here is where this example came from:

If $R$ is any ring and $\mathfrak m$ a maximal ideal in $R$, then the associated graded ring $gr(R)$ with respect to the $\mathfrak m$-adic filtration on $R$ is the same as the associated graded $gr(\widehat{R})$, where $\widehat{R}$ is the $\mathfrak m$-adic completion of $R$, again endowed with the $\mathfrak m$-adic filtration.

Now it can happen that $R$ is a domain but $\widehat{R}$ is not, e.g. because $R$ is the affine ring of an irreducible variety, which has more than one analytic branch passing through the point corresponding to $\mathfrak m$. The example I chose was perhaps the simplest one of this kind: $R$ is the affine ring of a nodal curve, and $\mathfrak m$ is the maximal ideal corresponding to the node, which has two analytic branches passing through it.

Added: This is identical to the example Steve just gave, and we were both led to it for the same reason.