The answer is "yes" for $n=2$ (that is the Shoenflies theorem). For $n>2$ the answer is "no". Consider a non-trivial knot $K$ in $\mathbb{R}^3$, and a continuous injective map $\partial\mathbb{D}^2\mapsto K$. Extend that map to a map $\mathbb{D}^2\to \mathbb{R}^3$. The resulting disk has a knot as a boundary, so it must have a self-intersection. As pointed out by M. Winter, that gives a counter-example for all $n\geq3$.

Another way to construct counter-examples is to consider an embedded disk with a single point of a self-intersection. For $n>=4$ one may consider a disk with a transversal self-intersection. For $n=3$ one may consider a disk that touches itself -- fold a piece of paper and pinch two halves together at one point.

The answer is "yes" for $n=2$ (that is the Schoenflies theorem). For $n>2$ the answer is "no". Consider a non-trivial knot $K$ in $\mathbb{R}^3$, and a continuous injective map $\partial\mathbb{D}^2\mapsto K$. Extend that map to a map $\mathbb{D}^2\to \mathbb{R}^3$. The resulting disk has a knot as a boundary, so it must have a self-intersection. As pointed out by M. Winter, that gives a counter-example for all $n\geq3$.

Another way to construct counter-examples is to consider an embedded disk with a single point of a self-intersection. For $n\ge 4$ one may consider a disk with a transversal self-intersection. For $n=3$ one may consider a disk that touches itself -- fold a piece of paper and pinch two halves together at one point.