$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\varepsilon} $ For any $x\in\R^d$, let us understand $\nabla f(x)$ as any one of the subgradients of the convex function $f$ at point $x$ (so, we will not have to worry about differentiability of $f$).

To obtain a contradiction, suppose that the desired conclusion does not hold. Then for some sequence $(x_n)$ in $\R^d$ such that $t_n:=|x_n|\to\infty$ and for some $\ell\in[-\infty,0]$ we have \begin{equation} u_n\cdot\nabla f(x_n)\to\ell \tag{1} \end{equation} where $u_n:=x_n/t_n$. By the compactness of the unit sphere $S^{d-1}$ in $\R^d$, without loss of generality (wlog) $u_n\to u$ for some $u\in S^{d-1}$.

For any $w\in\R^d$, consider the convex function $g_w\colon\R\to\R$ defined by the condition that $g_w(t)=f(tw)$ for all real $t$. Let $g'_w$ denote the left derivative of $g_w$. Then $g'_{u_n}(t_n)\le u_n\cdot\nabla f(x_n)$. So, in view of (1), wlog $g'_{u_n}(t_n)\to m$ for some $m\in[-\infty,\ell]\subseteq[-\infty,0]$.

So, for any $t\in[0,\infty)$ and any real $\ep>0$ there is some natural $n_{t,\ep}$ such that for all natural $n\ge n_{t,\ep}$ we have $t_n\ge t$ and $g'_{u_n}(t_n)\le\ep$, so that (by the convexity of $g_{u_n}$) $g'_{u_n}\le\ep$ on $[0,t_n]$ and hence on $[0,t]$, which yields $g_{u_n}(t)\le g_{u_n}(0)+\ep t$, that is, $f(tu_n)\le f(0)+\ep t$. Since $f$ is convex and real-valued, it is continuous. Therefore, $f(tu)\le f(0)+\ep t$, for all $t\in[0,\infty)$ and all real $\ep>0$. Thus, $f(tu)\le f(0)$ for all $t\in[0,\infty)$; that is, $f\le f(0)$ on $R_u:=\{tu\colon t\in[0,\infty)\}$.

Also, by the mentioned continuity of $f$, for some real $b$ we have $f\le b$ on $B:=\{x\in\R^d\colon|x|\le1\}$. So, $f\le \max[f(0),b]=b<\infty$ on the convex hull (say $C$) of $R_u\cup B$. The Lebesgue measure $|C|$ of $C$ is $\infty$. So, \begin{equation} \int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty, \end{equation} which contradicts the first display in the OP. $\Box$

$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\varepsilon} $ For any $x\in\R^d$, let us understand $\nabla f(x)$ as any one of the subgradients of the convex function $f$ at point $x$ (so, we will not have to worry about differentiability of $f$).

To obtain a contradiction, suppose that the desired conclusion does not hold. Then for some sequence $(x_n)$ in $\R^d$ such that $t_n:=|x_n|\to\infty$ and for some $\ell\in[-\infty,0]$ we have \begin{equation} u_n\cdot\nabla f(x_n)\to\ell \tag{1} \end{equation} where $u_n:=x_n/t_n$. By the compactness of the unit sphere $S^{d-1}$ in $\R^d$, without loss of generality (wlog) $u_n\to u$ for some $u\in S^{d-1}$.

For any $w\in\R^d$, consider the convex function $g_w\colon\R\to\R$ defined by the condition that $g_w(t)=f(tw)$ for all real $t$. Let $g'_w$ denote the left derivative of $g_w$. Then $g'_{u_n}(t_n)\le u_n\cdot\nabla f(x_n)$. So, in view of (1), wlog $g'_{u_n}(t_n)\to m$ for some $m\in[-\infty,\ell]\subseteq[-\infty,0]$.

So, for any $t\in[0,\infty)$ and any real $\ep>0$ there is some natural $n_{t,\ep}$ such that for all natural $n\ge n_{t,\ep}$ we have $t_n\ge t$ and $g'_{u_n}(t_n)\le\ep$, so that (by the convexity of $g_{u_n}$) $g'_{u_n}\le\ep$ on $[0,t_n]$ and hence on $[0,t]$, which yields $g_{u_n}(t)\le g_{u_n}(0)+\ep t$, that is, $f(tu_n)\le f(0)+\ep t$. Since $f$ is convex and real-valued, it is continuous. Therefore, $f(tu)\le f(0)+\ep t$, for all $t\in[0,\infty)$ and all real $\ep>0$. Thus, $f(tu)\le f(0)$ for all $t\in[0,\infty)$; that is, $f\le f(0)$ on $R_u:=\{tu\colon t\in[0,\infty)\}$.

Also, by the mentioned continuity of $f$, for some real $b$ we have $f\le b$ on $B:=\{x\in\R^d\colon|x|\le1\}$. So, $f\le \max[f(0),b]=b<\infty$ on the convex hull (say $C$) of $R_u\cup B$. The Lebesgue measure $|C|$ of $C$ is $\infty$. So, \begin{equation} \int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty, \end{equation} which contradicts the first display in the OP. $\Box$

Notes added in response to a comment by the OP:

1. The OP commented: "I'm not familiar with the concept of a subgradient. From Wikipedia, $\nabla f(x)$ is actually a set." Response: $\nabla f(x)$ was the notation you used, incorrectly. To make sense of such usage, I suggested: "let us understand $\nabla f(x)$ as any one of the subgradients".

2. The OP commented: "And we clearly need that each $\nabla f(x)$ is nonempty." Response: It is a well known fact that, for any real-valued convex function $f$ on $\R^d$ and any $x\in\R^d$, the subdiferential of $f$ at $x$ (which is the set of all subgradients of $f$ at $x$) is nonempty. See e.g. Theorem 23.4 or Proposition 2. Also, any real-valued convex function $f$ on $\R^d$ is continuous (and even locally Lipschitz continuous) -- see e.g. Theorem 10.1 or Theorem 3.3.1.

$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\varepsilon} $ For any $x\in\R^d$, let us understand $\nabla f(x)$ as any one of the subgradients of the convex function $f$ at point $x$ (so, we will not have to worry about differentiability of $f$).

To obtain a contradiction, suppose that the desired conclusion does not hold. Then for some sequence $(x_n)$ in $\R^d$ such that $t_n:=|x_n|\to\infty$ and for some $\ell\in[-\infty,0]$ we have \begin{equation} u_n\cdot\nabla f(x_n)\to\ell \tag{1} \end{equation} where $u_n:=x_n/t_n$. By the compactness of the unit sphere $S^{d-1}$ in $\R^d$, without loss of generality (wlog) $u_n\to u$ for some $u\in S^{d-1}$. For any $w\in\R^d$, consider the convex function $g_w\colon\R\to\R$ defined by the condition that $g_w(t)=f(tw)$ for all real $t$. Let $g'_w$ denote the left derivative of $g_w$. Then $g'_{u_n}(t_n)\le u_n\cdot\nabla f(x_n)$. So, in view of (1), wlog $g'_{u_n}(t_n)\to m$ for some $m\in[-\infty,\ell]\subseteq[-\infty,0]$.

So, for any $t\in[0,\infty)$ and any real $\ep>0$ there is some natural $n_{t,\ep}$ such that for all natural $n\ge n_{t,\ep}$ we have $t_n\ge t$ and $g'_{u_n}(t_n)\le\ep$, so that (by the convexity of $g_{u_n}$) $g'_{u_n}\le\ep$ on $[0,t_n]$ and hence on $[0,t]$, which yields $g_{u_n}(t)\le g_{u_n}(0)+\ep t$, that is, $f(tu_n)\le f(0)+\ep t$. Since $f$ is convex and real-valued, it is continuous. Therefore, $f(tu)\le f(0)+\ep t$, for all $t\in[0,\infty)$ and all real $\ep>0$. Thus, $f(tu)\le f(0)$ for all $t\in[0,\infty)$; that is, $f\le f(0)$ on $R_u:=\{tu\colon t\in[0,\infty)\}$.

Also, by the mentioned continuity of $f$, for some real $c$ we have $f\le c$ on $B:=\{x\in\R^d\colon|x|\le1\}$. So, $f\le b:=\max[c,f(0)]<\infty$ on the convex hull (say $C$) of $R_u\cup B$. The Lebesgue measure $|C|$ of $C$ is $\infty$. So, \begin{equation} \int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty, \end{equation} which contradicts the first display in the OP. $\Box$

$\newcommand{\R}{\mathbb{R}} \newcommand{\ep}{\varepsilon} $ For any $x\in\R^d$, let us understand $\nabla f(x)$ as any one of the subgradients of the convex function $f$ at point $x$ (so, we will not have to worry about differentiability of $f$).

To obtain a contradiction, suppose that the desired conclusion does not hold. Then for some sequence $(x_n)$ in $\R^d$ such that $t_n:=|x_n|\to\infty$ and for some $\ell\in[-\infty,0]$ we have \begin{equation} u_n\cdot\nabla f(x_n)\to\ell \tag{1} \end{equation} where $u_n:=x_n/t_n$. By the compactness of the unit sphere $S^{d-1}$ in $\R^d$, without loss of generality (wlog) $u_n\to u$ for some $u\in S^{d-1}$. 

For any $w\in\R^d$, consider the convex function $g_w\colon\R\to\R$ defined by the condition that $g_w(t)=f(tw)$ for all real $t$. Let $g'_w$ denote the left derivative of $g_w$. Then $g'_{u_n}(t_n)\le u_n\cdot\nabla f(x_n)$. So, in view of (1), wlog $g'_{u_n}(t_n)\to m$ for some $m\in[-\infty,\ell]\subseteq[-\infty,0]$.

So, for any $t\in[0,\infty)$ and any real $\ep>0$ there is some natural $n_{t,\ep}$ such that for all natural $n\ge n_{t,\ep}$ we have $t_n\ge t$ and $g'_{u_n}(t_n)\le\ep$, so that (by the convexity of $g_{u_n}$) $g'_{u_n}\le\ep$ on $[0,t_n]$ and hence on $[0,t]$, which yields $g_{u_n}(t)\le g_{u_n}(0)+\ep t$, that is, $f(tu_n)\le f(0)+\ep t$. Since $f$ is convex and real-valued, it is continuous. Therefore, $f(tu)\le f(0)+\ep t$, for all $t\in[0,\infty)$ and all real $\ep>0$. Thus, $f(tu)\le f(0)$ for all $t\in[0,\infty)$; that is, $f\le f(0)$ on $R_u:=\{tu\colon t\in[0,\infty)\}$.

Also, by the mentioned continuity of $f$, for some real $b$ we have $f\le b$ on $B:=\{x\in\R^d\colon|x|\le1\}$. So, $f\le \max[f(0),b]=b<\infty$ on the convex hull (say $C$) of $R_u\cup B$. The Lebesgue measure $|C|$ of $C$ is $\infty$. So, \begin{equation} \int e^{-f}\ge\int_C e^{-f}\ge|C|e^{-b}=\infty, \end{equation} which contradicts the first display in the OP. $\Box$