No. Consider the associated discrete valuation $v$ as a homomorphism $K((t))^\times \to \mathbb Z$. We have $K((t))^\times = K^\times \times t^{\mathbb Z} \times (1+ t K((t)))$. Elements of $1+t K((t))$ have arbitrarily high $n$th power roots in $K((t))$, hence they must be sent to $0$ by $v$. When restricted to $K$, $v$ must be a discrete valuation $v_0$ of $K$. Then we must have

$$ v( a_d t^d + a_{d+1} t^{d+1} +\dots ) = v_0(a_d) + c d $$ for some $c \in \mathbb Z$.

But valuations of this form clearly do not satisfy the inequality for the valuation of a sum unless $v_0$ is trivial, because we can make $v_0(a_d)$ very large and $v_0(a_{d+1})$ very small. So the only discrete valuation is the standard one $v( a_d t^d + a_{d+1} t^{d+1} +\dots ) =d$. But this has residue characteristic zero.

No. Consider the associated discrete valuation $v$ as a homomorphism $K((t))^\times \to \mathbb Z$. We have $K((t))^\times = K^\times \times t^{\mathbb Z} \times (1+ t K[[t]])$. Elements of $1+t K[[t]]$ have arbitrarily high $n$th power roots in $K((t))$, hence they must be sent to $0$ by $v$. When restricted to $K$, $v$ must be a discrete valuation $v_0$ of $K$. Then we must have

$$ v( a_d t^d + a_{d+1} t^{d+1} +\dots ) = v_0(a_d) + c d $$ for some $c \in \mathbb Z$.

But valuations of this form clearly do not satisfy the inequality for the valuation of a sum unless $v_0$ is trivial, because we can make $v_0(a_d)$ very large and $v_0(a_{d+1})$ very small. So the only discrete valuation is the standard one $v( a_d t^d + a_{d+1} t^{d+1} +\dots ) =d$. But this has residue characteristic zero.