Yes, and this doesn't require any assumption on $\mathcal C$. This follows from the following basic fact: if $I$ is filtered and $(A_i)_{i\in I}$ is a diagram of categories with colimit $A$, then the colimit of any functor $F: A\to\mathcal C$ may be computed as $$ colim_{a\in A} F(a) = colim_{i\in I}(colim_{a\in A_i}F_i(a)), $$ where $F_i$ is the restriction of $F$ to $A_i$ (provided the inner colimits all exist).
Write $\mathcal K=Ind_\lambda(\mathcal K_0)$. Every $X\in \mathcal K$ is then the colimit of the forgetful functor $\mathcal K_0/X \to \mathcal K_0\subset\mathcal K$, and $\mathcal K_0/X$ is $\lambda$-filtered. Let $I$ be $\lambda$-filtered and let $(X_i)_{i\in I}$ be a diagram in $\mathcal K$ with colimit $X$. Then $colim_i \mathcal K_0/X_i = \mathcal K_0/X$ and so $$ colim_{i} F(X_i) = colim_{i} (F(colim_{Y\to X_i} Y)) = colim_i (colim_{Y\to X_i} F(Y)) = colim_{Y \to X} F(Y) = F(colim_{Y\to X} Y)=F(X). $$