$\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then $$f(x)\sim\sqrt{\pi x}$$ (as $x\to\infty$).

Indeed, $$f(x)=\sum_{k\ge0}\frac{(1/2)_k}{k!}r_k(x)^2 =\sum_{k\ge0}2^{-2k}\binom{2k}k r_k(x)^2,$$ where $(a)_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and $$r_k(x):=\frac{(x)_k}{(x+1/2)_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)} \sim\sqrt{\frac x{x+k}}$$ uniformly in $k\ge0$ (as $x\to\infty$). So, $$f(x)\sim x\,\sum_{k\ge0}2^{-2k}\binom{2k}k \frac1{x+k} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)} \sim\sqrt{\pi x}$$ (as $x\to\infty$).

$\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then $$f(x)\sim\sqrt{\pi x}$$ (as $x\to\infty$).

Indeed, $$f(x)=\sum_{k\ge0}\frac{(1/2)_k}{k!}r_k(x)^2,$$ where $(a)_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and $$r_k(x):=\frac{(x)_k}{(x+1/2)_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)} \sim\sqrt{\frac x{x+k}}$$ uniformly in $k\ge0$ (as $x\to\infty$). So, \begin{equation*} f(x)\sim x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)} \sim\sqrt{\pi x} \tag{1}\label{1} \end{equation*} (as $x\to\infty$).

Details on the equality in \eqref{1}: \begin{equation*} \begin{aligned} \sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} & =\sum_{k\ge0}\frac{(1/2)_k}{k!} \int_0^1 dt\,t^{x+k-1} \\ & =\int_0^1 dt\,t^{x-1}\sum_{k\ge0}\frac{(1/2)_k}{k!} t^k \\ & =\int_0^1 dt\,t^{x-1}(1-t)^{-1/2}, \end{aligned} \end{equation*} in view of the Maclaurin series for $(1-t)^{-1/2}$. Therefore, \begin{equation*} \begin{aligned} x\,\sum_{k\ge0}\frac{(1/2)_k}{k!} \frac1{x+k} =x\,B(x,1/2)= x\,\frac{\Ga(1/2)\, \Ga(x)}{\Ga(x+1/2)} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}, \end{aligned} \end{equation*} as claimed.

Let $f(x)$ denote your hypergeometric expression. Then $$f(x)\sim\sqrt{\pi x}$$ as $x\to\infty$.

$\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then $$f(x)\sim\sqrt{\pi x}$$ (as $x\to\infty$).

Indeed, $$f(x)=\sum_{k\ge0}\frac{(1/2)_k}{k!}r_k(x)^2 =\sum_{k\ge0}2^{-2k}\binom{2k}k r_k(x)^2,$$ where $(a)_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and $$r_k(x):=\frac{(x)_k}{(x+1/2)_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)} \sim\sqrt{\frac x{x+k}}$$ uniformly in $k\ge0$ (as $x\to\infty$). So, $$f(x)\sim x\,\sum_{k\ge0}2^{-2k}\binom{2k}k \frac1{x+k} =\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)} \sim\sqrt{\pi x}$$ (as $x\to\infty$).