$\newcommand\ep\epsilon\newcommand\th\theta\newcommand\de\delta$The answer is yes, even with $C=1$ for all such $f$.
Indeed, let $f$ satisfy all your conditions on $f$. Let $g(0):=0$. For real $x>0$, let $$g(x):=1\wedge\inf\{f(\th_\de,\de)\colon\de\in E_x\},$$ where $1\wedge u:=\min(1,u)$ and $$E_x:=\{\de\in(0,1]\colon|\th_\de-\th_0|=x\}; $$ recall that $\inf\emptyset=\infty$.
Then $g(x)>0$ for all real $x>0$. Indeed, suppose the contrary: that $g(x)=0$ for some real $x>0$. Then $E_x\ne\emptyset$ and, moreover, there exist a sequence $(\de_n)$ in $E_x$ and some real $\ep\ge0$ such that $\de_n\to\ep$ and $f(\th_{\de_n},\de_n)\to0$ and $\th_{\de_n}\to\th$ for some $\th\ne\th_0$. So, then $$0=\lim_n f(\th_{\de_n},\de_n)=f(\th,0)>0,$$ since $\th\ne\th_0$. So, we get $0>0$, which proves that $g(x)>0$ for all real $x>0$.
Finally, take any $\ep\in(0,1]$. Then $\th_\ep\in E_{|\th_\ep-\th_0|}$ and hence $g(|\th_\ep-\th_0|)\le f(\th_\ep,\ep)$, which implies $$\limsup_{\ep\downarrow0}\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}\le1.\quad\Box$$
Replacing $g(x)$ by $xg(x)$, we can even get $$\lim_{\ep\downarrow0}\frac{g(|\th_\ep-\th_0|)}{f(\th_\ep,\ep)}=0.$$